{"id":776,"date":"2025-10-09T15:38:06","date_gmt":"2025-10-09T15:38:06","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=776"},"modified":"2025-11-20T17:22:35","modified_gmt":"2025-11-20T17:22:35","slug":"advanced-vector-calculus-background-youll-need-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/advanced-vector-calculus-background-youll-need-2\/","title":{"raw":"Advanced Vector Calculus: Background You'll Need 2","rendered":"Advanced Vector Calculus: Background You&#8217;ll Need 2"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li><span data-sheets-root=\"1\">Compute partial derivatives of multivariable functions<\/span><\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Derivatives of a Function of Two Variables<\/h2>\r\n<p class=\"whitespace-normal break-words\">In single-variable calculus, the derivative represents an instantaneous rate of change of [latex]y[\/latex] with respect to [latex]x[\/latex]. We write this using Leibniz notation as [latex]dy\/dx[\/latex], where [latex]y[\/latex] is the dependent variable and [latex]x[\/latex] is the independent variable.<\/p>\r\n<p class=\"whitespace-normal break-words\">For a function [latex]z=f(x,y)[\/latex] of two variables, both [latex]x[\/latex] and [latex]y[\/latex] are independent variables and [latex]z[\/latex] is the dependent variable. This immediately raises two questions: How do we adapt Leibniz notation for two variables? What does the derivative even mean when we have two inputs? The answer lies in <strong>partial derivatives<\/strong>.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>partial derivatives<\/h3>\r\nLet [latex]f(x, y)[\/latex] be a function fo two variables.\r\n\r\n[latex]\\\\[\/latex]\r\n<p class=\"whitespace-normal break-words\">The <strong>partial derivative of [latex]f[\/latex] with respect to [latex]x[\/latex]<\/strong>, written as [latex]\\frac{\\partial f}{\\partial x}[\/latex] or [latex]f_x[\/latex], is defined as<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\dfrac{\\partial f}{\\partial x}=\\underset{h\\to 0}{\\text{lim}}\\dfrac{f(x+h,y)-f(x,y)}{h}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">The <strong>partial derivative of [latex]f[\/latex] with respect to [latex]y[\/latex]<\/strong>, written as [latex]\\frac{\\partial f}{\\partial y}[\/latex] or [latex]f_y[\/latex], is defined as<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\dfrac{\\partial f}{\\partial y}=\\underset{k\\to 0}{\\text{lim}}\\dfrac{f(x,y+k)-f(x,y)}{k}[\/latex].<\/p>\r\n\r\n<\/section>\r\n<p class=\"whitespace-normal break-words\">This definition reveals two key differences from single-variable derivatives. First, the notation changes\u2014we use the symbol [latex]\\partial[\/latex] instead of [latex]d[\/latex]. This rounded \"[latex]d[\/latex]\" is called \"partial,\" so [latex]\\partial f\/\\partial x[\/latex] is read as \"the partial of [latex]f[\/latex] with respect to [latex]x[\/latex].\"<\/p>\r\n<p class=\"whitespace-normal break-words\">Second, we now have two different derivatives we can compute, one with respect to each independent variable. Depending on which variable we choose, we get entirely different partial derivatives.<\/p>\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">Use the definition of the partial derivative as a limit to calculate [latex]\\partial f\/\\partial x[\/latex] and [latex]\\partial f\/\\partial y[\/latex] for the function\r\n<p style=\"text-align: center;\">[latex]f(x, y)=x^{2}-3xy+2y^{2}-4x+5y-12[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1167793933114\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793933114\"]\r\n<p style=\"text-align: left;\">First, calculate [latex]f(x+h, y)[\/latex].<\/p>\r\n\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {f(x+h, y)} &amp; =\\hfill &amp; {(x+h)^{2}-3(x+h)y+2y^{2}-4(x+h)+5y-12} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {x^{2}+2xh+h^{2}-3xy-3hy+2y^{2}-4x-4h+5y-12.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nNext, substitute this into our definition and simplify:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{\\partial f}{\\partial x}} &amp; =\\hfill &amp; {\\displaystyle\\lim_{h\\to 0}\\frac{f(x+h, y)-f(x, y)}{h}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\lim_{h\\to 0}\\frac{(x^{2}+2xh+h^{2}-3xy-3hy+2y^{2}-4x-4h+5y-12)-(x^{2}-3xy+2y^{2}-4x+5y-12)}{h}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\lim_{h\\to 0}\\frac{x^{2}+2xh+h^{2}-3xy-3hy+2y^{2}-4x-4h+5y-12-x^{2}+3xy-2y^{2}+4x-5y+12}{h}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\lim_{h\\to 0}\\frac{2xh+h^{2}-3hy-4h}{h}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\lim_{h\\to 0}\\frac{h(2x+h-3y-4)}{h}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\lim_{h\\to 0}(2x+h-3y-4)}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {2x-3y-4}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nTo calculate [latex]\\frac{\\partial f}{\\partial y}[\/latex], first calculate [latex]f(x, y+h)[\/latex]:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {f(x, y+h)} &amp; =\\hfill &amp; {x^{2}-3x(y+h)+2(y+h)^{2}-4x+5(y+h)-12}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {x^{2}-3xy-3xh+2y^{2}+4yh+2h^{2}-4x+5y+5h-12.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nNext, substitute this into our definition and simplify:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{\\partial f}{\\partial y}} &amp; =\\hfill &amp; {\\displaystyle\\lim_{k\\to 0}\\frac{f(x, y+h)-f(x, y)}{k}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\lim_{k\\to 0}\\frac{(x^{2}-3xy-3xh+2y^{2}+4yh+2h^{2}-4x+5y+5h-12)-(x^{2}-3xy+2y^{2}-4x+5y-12)}{k}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\lim_{k\\to 0}\\frac{x^{2}-3xy-3xh+2y^{2}+4yh+2h^{2}-4x+5y+5h-12-x^{2}+3xy-2y^{2}+4x-5y+12}{k}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\lim_{k\\to 0}\\frac{-3xk+4yk+2k^{2}+5k}{k}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\lim_{k\\to 0}\\frac{h(-3x+4y+2k+5)}{k}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\lim_{k\\to 0}(-3x+4y+2k+5)}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {-3x+4y+5.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<h3 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">Computing Partial Derivatives<\/h3>\r\n<p class=\"whitespace-normal break-words\">The key idea when calculating partial derivatives is simple: treat all independent variables except the one you're differentiating with respect to as constants. Then differentiate as you would with a single-variable function.<\/p>\r\n<p class=\"whitespace-normal break-words\">Why does this work? When we fix [latex]y[\/latex] and define [latex]g(x)=f(x,y)[\/latex] as a function of [latex]x[\/latex] only, we get:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]g'(x)=\\underset{h\\to 0}{\\text{lim}}\\dfrac{g(x+h)-g(x)}{h}=\\underset{h\\to 0}{\\text{lim}}\\dfrac{f(x+h,y)-f(x,y)}{h}=\\dfrac{\\partial f}{\\partial x}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">The same reasoning applies for the partial derivative with respect to [latex]y[\/latex]. Fix [latex]x[\/latex] and define [latex]h(y)=f(x,y)[\/latex] as a function of [latex]y[\/latex] only:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]h'(y)=\\underset{k\\to 0}{\\text{lim}}\\dfrac{h(y+k)-h(y)}{k}=\\underset{k\\to 0}{\\text{lim}}\\dfrac{f(x,y+k)-f(x,y)}{k}=\\dfrac{\\partial f}{\\partial y}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">All the differentiation rules you learned in single-variable calculus still apply.<\/p>\r\n\r\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">\r\n<p class=\"whitespace-normal break-words\">To find [latex]\\dfrac{\\partial f}{\\partial x}[\/latex]: Treat [latex]y[\/latex] as a constant and differentiate with respect to [latex]x[\/latex].<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">To find [latex]\\dfrac{\\partial f}{\\partial y}[\/latex]: Treat [latex]x[\/latex] as a constant and differentiate with respect to [latex]y[\/latex].<\/p>\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Calculate [latex]\\partial f\/\\partial x[\/latex] and [latex]\\partial f\/\\partial y[\/latex] for the following functions by holding the opposite variable constant then differentiating:\r\n<ol>\r\n \t<li style=\"text-align: left;\">[latex]f(x, y)=x^{2}-3xy+2y^{2}-4x+5y-12[\/latex].<\/li>\r\n \t<li>[latex]g(x, y)=\\sin{(x^{2}y-2x+4)}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1267793933114\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1267793933114\"]\r\n<ol>\r\n \t<li style=\"text-align: left;\">To calculate [latex]\\partial f\/\\partial x[\/latex], treat the variable [latex]y[\/latex] as a constant. Then differentiate [latex]f(x, y)[\/latex] with respect to [latex]x[\/latex] using the sum, difference, and power rules:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{\\partial f}{\\partial x}} &amp; =\\hfill &amp; {\\frac{\\partial}{\\partial x}\\big[x^{2}-3xy+2y^{2}-4x+5y-12\\big]}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{\\partial}{\\partial x}\\big[x^{2}\\big]-\\frac{\\partial}{\\partial x}\\big[3xy\\big]+\\frac{\\partial}{\\partial x}\\big[2y^{2}\\big]-\\frac{\\partial}{\\partial x}\\big[4x\\big]+\\frac{\\partial}{\\partial x}\\big[5y\\big]-\\frac{\\partial}{\\partial x}\\big[12\\big]}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {2x-3y+0-4+0-0}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {2x-3y-4.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nThe derivatives of the third, fifth, and sixth terms are all zero because they do not contain the variable [latex]x[\/latex], so they are treated as constant terms. The derivative of the second term is equal to the coefficient of [latex]x[\/latex], which is [latex]-3y[\/latex]. Calculating [latex]\\partial f\/\\partial y[\/latex]:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{\\partial f}{\\partial y}} &amp; =\\hfill &amp; {\\frac{\\partial}{\\partial y}\\big[x^{2}-3xy+2y^{2}-4x+5y-12\\big]}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{\\partial}{\\partial y}\\big[x^{2}\\big]-\\frac{\\partial}{\\partial y}\\big[3xy\\big]+\\frac{\\partial}{\\partial y}\\big[2y^{2}\\big]-\\frac{\\partial}{\\partial y}\\big[4x\\big]+\\frac{\\partial}{\\partial y}\\big[5y\\big]-\\frac{\\partial}{\\partial y}\\big[12\\big]}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {-3x+4y-0+5-0}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {-3x+4y+5.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nThese are the same answers obtained in the previous example<\/li>\r\n \t<li>To calculate [latex]\\partial g\/\\partial x[\/latex], create the variable [latex]y[\/latex] as a constant. Then differentiate [latex]g(x, y)[\/latex] with respect to [latex]x[\/latex] using the chain rule and power rule:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill{\\frac{\\partial g}{\\partial y}} &amp; =\\hfill &amp; {\\frac{\\partial}{\\partial x}\\big[\\sin{(x^{2}y-2x+4)}\\big]} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\cos{(x^{2}y-2x+4)}\\frac{\\partial}{\\partial x}[x^{2}y-2x+4]}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {(2xy-2)\\cos{(x^{2}y-2x+4)}.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nTo calculate [latex]\\partial f\/\\partial y[\/latex], treat the variable [latex]x[\/latex] as a constant. Then differentiate [latex]g(x, y)[\/latex] with respect to [latex]y[\/latex] using the chain rule and power rule:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{\\partial g}{\\partial y}} &amp; =\\hfill &amp; {\\frac{\\partial}{\\partial y}\\big[\\sin{(x^{2}y-2x+4)}\\big]}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\cos{(x^{2}y-2x+4)}\\frac{\\partial}{\\partial x}[x^{2}y-2x+4]} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {x^{2}\\cos{(x^{2}-2x+4)}.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]6041[\/ohm_question]<\/section>\r\n<h2>Functions of More Than Two Variables<\/h2>\r\nSuppose we have a function of three variables, such as [latex]w=f(x,y,z)[\/latex]. We can calculate partial derivatives of [latex]w[\/latex] with respect to any of the independent variables. The definitions extend naturally from what we did with two variables.\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>partial derivatives of functions of three variables<\/h3>\r\n<p class=\"whitespace-normal break-words\">Let [latex]f(x,y,z)[\/latex] be a function of three variables. Then:<\/p>\r\n[latex]\\\\[\/latex]\r\n<p class=\"whitespace-normal break-words\">The <strong>partial derivative with respect to [latex]x[\/latex]<\/strong> is<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\dfrac{\\partial f}{\\partial x}=\\underset{h\\to 0}{\\text{lim}}\\dfrac{f(x+h,y,z)-f(x,y,z)}{h}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">The <strong>partial derivative with respect to [latex]y[\/latex]<\/strong> is<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\dfrac{\\partial f}{\\partial y}=\\underset{k\\to 0}{\\text{lim}}\\dfrac{f(x,y+k,z)-f(x,y,z)}{k}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">The <strong>partial derivative with respect to [latex]z[\/latex]<\/strong> is<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\dfrac{\\partial f}{\\partial z}=\\underset{m\\to 0}{\\text{lim}}\\dfrac{f(x,y,z+m)-f(x,y,z)}{m}[\/latex].<\/p>\r\n\r\n<\/section>The computation technique is identical to what you used with two variables. To calculate [latex]\\frac{\\partial f}{\\partial x}[\/latex], treat [latex]y[\/latex] and [latex]z[\/latex] as constants and differentiate with respect to [latex]x[\/latex]. The same principle applies for the other partial derivatives\u2014just hold the other variables constant and differentiate with respect to the variable of interest.\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">Use the limit definition of partial derivatives to calculate [latex]\\partial f\/\\partial x[\/latex] for the function\r\n<div style=\"text-align: center;\">[latex]\\large{f\\,(x,\\ y,\\ z)=x^{2}-3xy+2y^{2}-4xz+5yz^{2}-12x+4y-3z}[\/latex].<\/div>\r\n&nbsp;\r\n\r\nThen, find [latex]\\partial f\/\\partial y[\/latex] and [latex]\\partial f\/\\partial z[\/latex] by setting the other two variables constant and differentiating accordingly.\r\n\r\n[reveal-answer q=\"fs-id1367903938124\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1367903938124\"]\r\n<p style=\"text-align: left;\">We first calculate [latex]\\partial f\/\\partial x[\/latex] using our definition, then we calculate the other two partial derivatives by holding the remaining variables constant. To use the equation to find [latex]\\partial f\/\\partial x[\/latex], we first need to calculate [latex]f\\,(x+h,\\ y,\\ z)[\/latex]:<\/p>\r\n\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {f\\,(x+h,\\ y,\\ z)} &amp; =\\hfill &amp; {(x+h)^{2}-3(x+h)y+2y^{2}-4(x+h)z+5yz^{2}-12(x+h)+4y-3z}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {x^{2}+2xh+h^{2}-3xy-3xh+2y^{2}-4xz-4hz+5yz^{2}-12x-12h+4y-3z} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nand recall that [latex]f\\,(x,\\ y,\\ z)=x^{2}-3xy+2y^{2}-4xz+5yz^{2}-12x+4y-3z[\/latex]. Next, we substitute these two expressions into the equation:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{\\partial f}{\\partial x}} &amp; =\\hfill &amp; {\\displaystyle\\lim_{h\\to 0}\\left[\\frac{x^{2}+2xh+h^{2}-3xy-3xh+2y^{2}-4xz-4hz+5yz^{2}-12x-12h+4y-3z}{h} \\\\ - \\frac{x^{2}-3xy+2y^{2}-4xz+5yz^{2}-12x+4y-3z}{h}\\right]}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\lim_{h\\to 0}\\left[\\frac{2xh+h^{2}-3hy-4hz-12h}{h}\\right]}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\lim_{h\\to 0}\\left[\\frac{h(2x+h-3y-4z-12)}{h}\\right]}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\lim_{h\\to 0}(2x+h-3y-4z-12)}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {2x-3y-4z-12.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nThen we find [latex]\\partial f\/\\partial y[\/latex] by holding [latex]x[\/latex] and [latex]z[\/latex] constant. Therefore, any term that does not include the variable [latex]y[\/latex] is constant, and its derivative is zero. We can apply the sum, difference, and power rules for functions of one variable:\r\n\r\n[latex] \\begin{alignat}{2} &amp;\\frac{\\partial}{\\partial y}\\left[x^{2}-3xy+2y^{2}-4xz+5yz^{2}-12x+4y-3z\\right] \\\\\r\n\r\n\\hspace{2cm}&amp;=\\frac{\\partial}{\\partial y}\\left[x^{2}\\right]-\\frac{\\partial}{\\partial y}\\left[3xy\\right]+\\frac{\\partial}{\\partial y}\\left[2y^{2}\\right]-\\frac{\\partial}{\\partial y}\\left[4xz\\right]+\\frac{\\partial}{\\partial y}\\left[5yz^{2}\\right]-\\frac{\\partial}{\\partial y}\\left[12x\\right]+\\frac{\\partial}{\\partial y}\\left[4y\\right]-\\frac{\\partial}{\\partial y}\\left[3z\\right] &amp;\\quad\\\\\r\n\r\n&amp;=0-3x+4y-0+5z^2-0+4-0\\\\\r\n\r\n&amp;=-3x+4y+5z^2+4.\\\\\r\n\r\n\\end{alignat}[\/latex]\r\n\r\nTo calculate\u00a0[latex]\\partial f\/\\partial z[\/latex], we hold [latex]x[\/latex] and [latex]y[\/latex] constant and apply the sum, difference, and power rules for functions of one variable:\r\n\r\n[latex] \\begin{alignat}{2} &amp;\\frac{\\partial}{\\partial z}\\left[x^{2}-3xy+2y^{2}-4xz+5yz^{2}-12x+4y-3z\\right] \\\\\r\n\r\n\\hspace{2cm}&amp;=\\frac{\\partial}{\\partial z}\\left[x^{2}\\right]-\\frac{\\partial}{\\partial z}\\left[3xy\\right]+\\frac{\\partial}{\\partial z}\\left[2y^{2}\\right]-\\frac{\\partial}{\\partial z}\\left[4xz\\right]+\\frac{\\partial}{\\partial z}\\left[5yz^{2}\\right]-\\frac{\\partial}{\\partial z}\\left[12x\\right]+\\frac{\\partial}{\\partial z}\\left[4y\\right]-\\frac{\\partial}{\\partial y}\\left[3z\\right] &amp;\\quad\\\\\r\n\r\n&amp;=0-0+0-4x+10yz-0+0-3\\\\\r\n\r\n&amp;=-4z+10yz-3.\\\\\r\n\r\n\\end{alignat}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Calculate the three partial derivatives of the following functions.\r\n<ol>\r\n \t<li>[latex]f\\,(x,\\ y,\\ z)=\\frac{x^{2}y-4xz+y^{2}}{x-3yz}[\/latex]<\/li>\r\n \t<li>[latex]g\\,(x,\\ y,\\ z)=\\sin{(x^{2}y-z)}+\\cos{(x^{2}-yz)}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1367703960124\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1367703960124\"]\r\n<div style=\"text-align: left;\">In each case, treat all variables as constants except the one whose partial derivative you are calculating.<\/div>\r\n&nbsp;\r\n<p style=\"padding-left: 30px;\">[latex]\\begin{array}{ccc}1. \\hspace{1cm} {\\frac{\\partial f}{\\partial x}} &amp; =\\hfill &amp; {\\frac{\\partial}{\\partial x}\\bigg[\\frac{x^{2}y-4xz+y^{2}}{x-3yz}\\bigg]}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{\\frac{\\partial}{\\partial x}(x^{2}y-4xz+y^{2})(x-3yz)-(x^{2}y-4xz+y^{2})\\frac{\\partial}{\\partial x}(x-3yz)}{(x-3yz)^{2}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{(2xy-4z)(x-3yz)-(x^{2}y-4xz+y^{2})(1)}{(x-3yz)^{2}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{2x^{2}y-6xy^{2}z-4xz+12yz^{2}-x^{2}y+4xz-y^{2}}{(x-3yz)^{2}}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{x^{2}y-6xy^{2}z-4xz+12yz^{2}+4xz-y^{2}}{(x-3yz)^{2}}}\\hfill \\\\ \\hfill {\\frac{\\partial f}{\\partial y}} &amp; =\\hfill &amp; {\\frac{\\partial}{\\partial y}\\bigg[\\frac{x^{2}y-4xz+y^{2}}{x-3yz}\\bigg]}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{\\frac{\\partial}{\\partial y}(x^{2}y-4xz+y^{2})(x-3yz)-(x^{2}y-4xz+y^{2})\\frac{\\partial}{\\partial y}(x-3yz)}{(x-3yz)^{2}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{(x^{2}+2y)(x-3yz)-(x^{2}y-4xz+y^{2})(-3z)}{(x-3yz)^{2}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{x^{3}-3x^{2}yz+2xy-6y^{2}z+3x^{2}yz-12xz^{2}+3y^{2}z}{(x-3yz)^{2}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{x^{3}+2xy-3y^{2}z-12xz^{2}}{(x-3yz)^{2}}}\\hfill \\\\ \\hfill {\\frac{\\partial f}{\\partial z}} &amp; =\\hfill &amp; {\\frac{\\partial}{\\partial z}\\bigg[\\frac{x^{2}y-4xz+y^{2}}{x-3yz}\\bigg]}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{\\frac{\\partial}{\\partial z}(x^{2}y-4xz+y^{2})(x-3yz)-(x^{2}y-4xz+y^{2})\\frac{\\partial}{\\partial z}(x-3yz)}{(x-3yz)^{2}}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{(-4x)(x-3yz)-(x^{2}y-4xz+y^{2})(-3y)}{(x-3yz)^{2}}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{-4x^{2}+12xyz+3x^{2}y^{2}-12xyz+3y^{3}}{(x-3yz)^{2}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{-4x^{2}+3x^{2}y^{2}+3y^{3}}{(x-3yz)^{2}}} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/p>\r\n<p style=\"padding-left: 30px;\">[latex]\\begin{array}{ccc}2. \\hspace{1cm} {\\frac{\\partial f}{\\partial x}} &amp; =\\hfill &amp; {\\frac{\\partial}{\\partial x}\\left[\\sin{(x^{2}y-z)}+\\cos{(x^{2}-yz)}\\right]}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {(\\cos{(x^{2}y-z)})\\frac{\\partial}{\\partial x}(x^{2}y-z)-(\\sin{(x^{2}-yz)})\\frac{\\partial}{\\partial x}(x^{2}-yz)}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {2xy\\cos{(x^{2}y-z)}-2x\\sin{(x^{2}-yz)}} \\hfill \\\\ \\hfill {\\frac{\\partial f}{\\partial y}} &amp; =\\hfill &amp; {\\frac{\\partial}{\\partial y}\\left[\\sin{(x^{2}y-z)}+\\cos{(x^{2}-yz)}\\right]}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {(\\cos{(x^{2}y-z)})\\frac{\\partial}{\\partial y}(x^{2}y-z)-(\\sin{(x^{2}-yz)})\\frac{\\partial}{\\partial y}(x^{2}-yz)}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {x^{2}\\cos{(x^{2}y-z)}+z\\sin{(x^{2}-yz)}} \\hfill \\\\ \\hfill {\\frac{\\partial f}{\\partial z}} &amp; =\\hfill &amp; {\\frac{\\partial}{\\partial z}\\left[\\sin{(x^{2}y-z)}+\\cos{(x^{2}-yz)}\\right]}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {(\\cos{(x^{2}y-z)})\\frac{\\partial}{\\partial z}(x^{2}y-z)-(\\sin{(x^{2}-yz)})\\frac{\\partial}{\\partial z}(x^{2}-yz)} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {-\\cos{(x^{2}y-z)}+y\\sin{(x^{2}-yz)}} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li><span data-sheets-root=\"1\">Compute partial derivatives of multivariable functions<\/span><\/li>\n<\/ul>\n<\/section>\n<h2>Derivatives of a Function of Two Variables<\/h2>\n<p class=\"whitespace-normal break-words\">In single-variable calculus, the derivative represents an instantaneous rate of change of [latex]y[\/latex] with respect to [latex]x[\/latex]. We write this using Leibniz notation as [latex]dy\/dx[\/latex], where [latex]y[\/latex] is the dependent variable and [latex]x[\/latex] is the independent variable.<\/p>\n<p class=\"whitespace-normal break-words\">For a function [latex]z=f(x,y)[\/latex] of two variables, both [latex]x[\/latex] and [latex]y[\/latex] are independent variables and [latex]z[\/latex] is the dependent variable. This immediately raises two questions: How do we adapt Leibniz notation for two variables? What does the derivative even mean when we have two inputs? The answer lies in <strong>partial derivatives<\/strong>.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>partial derivatives<\/h3>\n<p>Let [latex]f(x, y)[\/latex] be a function fo two variables.<\/p>\n<p>[latex]\\\\[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">The <strong>partial derivative of [latex]f[\/latex] with respect to [latex]x[\/latex]<\/strong>, written as [latex]\\frac{\\partial f}{\\partial x}[\/latex] or [latex]f_x[\/latex], is defined as<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\dfrac{\\partial f}{\\partial x}=\\underset{h\\to 0}{\\text{lim}}\\dfrac{f(x+h,y)-f(x,y)}{h}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">The <strong>partial derivative of [latex]f[\/latex] with respect to [latex]y[\/latex]<\/strong>, written as [latex]\\frac{\\partial f}{\\partial y}[\/latex] or [latex]f_y[\/latex], is defined as<\/p>\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\dfrac{\\partial f}{\\partial y}=\\underset{k\\to 0}{\\text{lim}}\\dfrac{f(x,y+k)-f(x,y)}{k}[\/latex].<\/p>\n<\/section>\n<p class=\"whitespace-normal break-words\">This definition reveals two key differences from single-variable derivatives. First, the notation changes\u2014we use the symbol [latex]\\partial[\/latex] instead of [latex]d[\/latex]. This rounded &#8220;[latex]d[\/latex]&#8221; is called &#8220;partial,&#8221; so [latex]\\partial f\/\\partial x[\/latex] is read as &#8220;the partial of [latex]f[\/latex] with respect to [latex]x[\/latex].&#8221;<\/p>\n<p class=\"whitespace-normal break-words\">Second, we now have two different derivatives we can compute, one with respect to each independent variable. Depending on which variable we choose, we get entirely different partial derivatives.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">Use the definition of the partial derivative as a limit to calculate [latex]\\partial f\/\\partial x[\/latex] and [latex]\\partial f\/\\partial y[\/latex] for the function<\/p>\n<p style=\"text-align: center;\">[latex]f(x, y)=x^{2}-3xy+2y^{2}-4x+5y-12[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167793933114\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167793933114\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: left;\">First, calculate [latex]f(x+h, y)[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {f(x+h, y)} & =\\hfill & {(x+h)^{2}-3(x+h)y+2y^{2}-4(x+h)+5y-12} \\hfill \\\\ \\hfill & =\\hfill & {x^{2}+2xh+h^{2}-3xy-3hy+2y^{2}-4x-4h+5y-12.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Next, substitute this into our definition and simplify:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{\\partial f}{\\partial x}} & =\\hfill & {\\displaystyle\\lim_{h\\to 0}\\frac{f(x+h, y)-f(x, y)}{h}} \\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{h\\to 0}\\frac{(x^{2}+2xh+h^{2}-3xy-3hy+2y^{2}-4x-4h+5y-12)-(x^{2}-3xy+2y^{2}-4x+5y-12)}{h}} \\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{h\\to 0}\\frac{x^{2}+2xh+h^{2}-3xy-3hy+2y^{2}-4x-4h+5y-12-x^{2}+3xy-2y^{2}+4x-5y+12}{h}}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{h\\to 0}\\frac{2xh+h^{2}-3hy-4h}{h}}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{h\\to 0}\\frac{h(2x+h-3y-4)}{h}}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{h\\to 0}(2x+h-3y-4)}\\hfill \\\\ \\hfill & =\\hfill & {2x-3y-4}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>To calculate [latex]\\frac{\\partial f}{\\partial y}[\/latex], first calculate [latex]f(x, y+h)[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {f(x, y+h)} & =\\hfill & {x^{2}-3x(y+h)+2(y+h)^{2}-4x+5(y+h)-12}\\hfill \\\\ \\hfill & =\\hfill & {x^{2}-3xy-3xh+2y^{2}+4yh+2h^{2}-4x+5y+5h-12.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Next, substitute this into our definition and simplify:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{\\partial f}{\\partial y}} & =\\hfill & {\\displaystyle\\lim_{k\\to 0}\\frac{f(x, y+h)-f(x, y)}{k}}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{k\\to 0}\\frac{(x^{2}-3xy-3xh+2y^{2}+4yh+2h^{2}-4x+5y+5h-12)-(x^{2}-3xy+2y^{2}-4x+5y-12)}{k}}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{k\\to 0}\\frac{x^{2}-3xy-3xh+2y^{2}+4yh+2h^{2}-4x+5y+5h-12-x^{2}+3xy-2y^{2}+4x-5y+12}{k}}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{k\\to 0}\\frac{-3xk+4yk+2k^{2}+5k}{k}}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{k\\to 0}\\frac{h(-3x+4y+2k+5)}{k}}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{k\\to 0}(-3x+4y+2k+5)}\\hfill \\\\ \\hfill & =\\hfill & {-3x+4y+5.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<h3 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">Computing Partial Derivatives<\/h3>\n<p class=\"whitespace-normal break-words\">The key idea when calculating partial derivatives is simple: treat all independent variables except the one you&#8217;re differentiating with respect to as constants. Then differentiate as you would with a single-variable function.<\/p>\n<p class=\"whitespace-normal break-words\">Why does this work? When we fix [latex]y[\/latex] and define [latex]g(x)=f(x,y)[\/latex] as a function of [latex]x[\/latex] only, we get:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]g'(x)=\\underset{h\\to 0}{\\text{lim}}\\dfrac{g(x+h)-g(x)}{h}=\\underset{h\\to 0}{\\text{lim}}\\dfrac{f(x+h,y)-f(x,y)}{h}=\\dfrac{\\partial f}{\\partial x}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">The same reasoning applies for the partial derivative with respect to [latex]y[\/latex]. Fix [latex]x[\/latex] and define [latex]h(y)=f(x,y)[\/latex] as a function of [latex]y[\/latex] only:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]h'(y)=\\underset{k\\to 0}{\\text{lim}}\\dfrac{h(y+k)-h(y)}{k}=\\underset{k\\to 0}{\\text{lim}}\\dfrac{f(x,y+k)-f(x,y)}{k}=\\dfrac{\\partial f}{\\partial y}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">All the differentiation rules you learned in single-variable calculus still apply.<\/p>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">\n<p class=\"whitespace-normal break-words\">To find [latex]\\dfrac{\\partial f}{\\partial x}[\/latex]: Treat [latex]y[\/latex] as a constant and differentiate with respect to [latex]x[\/latex].<\/p>\n<p class=\"whitespace-pre-wrap break-words\">To find [latex]\\dfrac{\\partial f}{\\partial y}[\/latex]: Treat [latex]x[\/latex] as a constant and differentiate with respect to [latex]y[\/latex].<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Calculate [latex]\\partial f\/\\partial x[\/latex] and [latex]\\partial f\/\\partial y[\/latex] for the following functions by holding the opposite variable constant then differentiating:<\/p>\n<ol>\n<li style=\"text-align: left;\">[latex]f(x, y)=x^{2}-3xy+2y^{2}-4x+5y-12[\/latex].<\/li>\n<li>[latex]g(x, y)=\\sin{(x^{2}y-2x+4)}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1267793933114\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1267793933114\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li style=\"text-align: left;\">To calculate [latex]\\partial f\/\\partial x[\/latex], treat the variable [latex]y[\/latex] as a constant. Then differentiate [latex]f(x, y)[\/latex] with respect to [latex]x[\/latex] using the sum, difference, and power rules:\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{\\partial f}{\\partial x}} & =\\hfill & {\\frac{\\partial}{\\partial x}\\big[x^{2}-3xy+2y^{2}-4x+5y-12\\big]}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{\\partial}{\\partial x}\\big[x^{2}\\big]-\\frac{\\partial}{\\partial x}\\big[3xy\\big]+\\frac{\\partial}{\\partial x}\\big[2y^{2}\\big]-\\frac{\\partial}{\\partial x}\\big[4x\\big]+\\frac{\\partial}{\\partial x}\\big[5y\\big]-\\frac{\\partial}{\\partial x}\\big[12\\big]}\\hfill \\\\ \\hfill & =\\hfill & {2x-3y+0-4+0-0}\\hfill \\\\ \\hfill & =\\hfill & {2x-3y-4.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>The derivatives of the third, fifth, and sixth terms are all zero because they do not contain the variable [latex]x[\/latex], so they are treated as constant terms. The derivative of the second term is equal to the coefficient of [latex]x[\/latex], which is [latex]-3y[\/latex]. Calculating [latex]\\partial f\/\\partial y[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{\\partial f}{\\partial y}} & =\\hfill & {\\frac{\\partial}{\\partial y}\\big[x^{2}-3xy+2y^{2}-4x+5y-12\\big]}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{\\partial}{\\partial y}\\big[x^{2}\\big]-\\frac{\\partial}{\\partial y}\\big[3xy\\big]+\\frac{\\partial}{\\partial y}\\big[2y^{2}\\big]-\\frac{\\partial}{\\partial y}\\big[4x\\big]+\\frac{\\partial}{\\partial y}\\big[5y\\big]-\\frac{\\partial}{\\partial y}\\big[12\\big]}\\hfill \\\\ \\hfill & =\\hfill & {-3x+4y-0+5-0}\\hfill \\\\ \\hfill & =\\hfill & {-3x+4y+5.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>These are the same answers obtained in the previous example<\/li>\n<li>To calculate [latex]\\partial g\/\\partial x[\/latex], create the variable [latex]y[\/latex] as a constant. Then differentiate [latex]g(x, y)[\/latex] with respect to [latex]x[\/latex] using the chain rule and power rule:\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill{\\frac{\\partial g}{\\partial y}} & =\\hfill & {\\frac{\\partial}{\\partial x}\\big[\\sin{(x^{2}y-2x+4)}\\big]} \\hfill \\\\ \\hfill & =\\hfill & {\\cos{(x^{2}y-2x+4)}\\frac{\\partial}{\\partial x}[x^{2}y-2x+4]}\\hfill \\\\ \\hfill & =\\hfill & {(2xy-2)\\cos{(x^{2}y-2x+4)}.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>To calculate [latex]\\partial f\/\\partial y[\/latex], treat the variable [latex]x[\/latex] as a constant. Then differentiate [latex]g(x, y)[\/latex] with respect to [latex]y[\/latex] using the chain rule and power rule:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{\\partial g}{\\partial y}} & =\\hfill & {\\frac{\\partial}{\\partial y}\\big[\\sin{(x^{2}y-2x+4)}\\big]}\\hfill \\\\ \\hfill & =\\hfill & {\\cos{(x^{2}y-2x+4)}\\frac{\\partial}{\\partial x}[x^{2}y-2x+4]} \\hfill \\\\ \\hfill & =\\hfill & {x^{2}\\cos{(x^{2}-2x+4)}.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm6041\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=6041&theme=lumen&iframe_resize_id=ohm6041&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<h2>Functions of More Than Two Variables<\/h2>\n<p>Suppose we have a function of three variables, such as [latex]w=f(x,y,z)[\/latex]. We can calculate partial derivatives of [latex]w[\/latex] with respect to any of the independent variables. The definitions extend naturally from what we did with two variables.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>partial derivatives of functions of three variables<\/h3>\n<p class=\"whitespace-normal break-words\">Let [latex]f(x,y,z)[\/latex] be a function of three variables. Then:<\/p>\n<p>[latex]\\\\[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">The <strong>partial derivative with respect to [latex]x[\/latex]<\/strong> is<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\dfrac{\\partial f}{\\partial x}=\\underset{h\\to 0}{\\text{lim}}\\dfrac{f(x+h,y,z)-f(x,y,z)}{h}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">The <strong>partial derivative with respect to [latex]y[\/latex]<\/strong> is<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\dfrac{\\partial f}{\\partial y}=\\underset{k\\to 0}{\\text{lim}}\\dfrac{f(x,y+k,z)-f(x,y,z)}{k}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">The <strong>partial derivative with respect to [latex]z[\/latex]<\/strong> is<\/p>\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\dfrac{\\partial f}{\\partial z}=\\underset{m\\to 0}{\\text{lim}}\\dfrac{f(x,y,z+m)-f(x,y,z)}{m}[\/latex].<\/p>\n<\/section>\n<p>The computation technique is identical to what you used with two variables. To calculate [latex]\\frac{\\partial f}{\\partial x}[\/latex], treat [latex]y[\/latex] and [latex]z[\/latex] as constants and differentiate with respect to [latex]x[\/latex]. The same principle applies for the other partial derivatives\u2014just hold the other variables constant and differentiate with respect to the variable of interest.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">Use the limit definition of partial derivatives to calculate [latex]\\partial f\/\\partial x[\/latex] for the function<\/p>\n<div style=\"text-align: center;\">[latex]\\large{f\\,(x,\\ y,\\ z)=x^{2}-3xy+2y^{2}-4xz+5yz^{2}-12x+4y-3z}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p>Then, find [latex]\\partial f\/\\partial y[\/latex] and [latex]\\partial f\/\\partial z[\/latex] by setting the other two variables constant and differentiating accordingly.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1367903938124\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1367903938124\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: left;\">We first calculate [latex]\\partial f\/\\partial x[\/latex] using our definition, then we calculate the other two partial derivatives by holding the remaining variables constant. To use the equation to find [latex]\\partial f\/\\partial x[\/latex], we first need to calculate [latex]f\\,(x+h,\\ y,\\ z)[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {f\\,(x+h,\\ y,\\ z)} & =\\hfill & {(x+h)^{2}-3(x+h)y+2y^{2}-4(x+h)z+5yz^{2}-12(x+h)+4y-3z}\\hfill \\\\ \\hfill & =\\hfill & {x^{2}+2xh+h^{2}-3xy-3xh+2y^{2}-4xz-4hz+5yz^{2}-12x-12h+4y-3z} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>and recall that [latex]f\\,(x,\\ y,\\ z)=x^{2}-3xy+2y^{2}-4xz+5yz^{2}-12x+4y-3z[\/latex]. Next, we substitute these two expressions into the equation:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{\\partial f}{\\partial x}} & =\\hfill & {\\displaystyle\\lim_{h\\to 0}\\left[\\frac{x^{2}+2xh+h^{2}-3xy-3xh+2y^{2}-4xz-4hz+5yz^{2}-12x-12h+4y-3z}{h} \\\\ - \\frac{x^{2}-3xy+2y^{2}-4xz+5yz^{2}-12x+4y-3z}{h}\\right]}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{h\\to 0}\\left[\\frac{2xh+h^{2}-3hy-4hz-12h}{h}\\right]}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{h\\to 0}\\left[\\frac{h(2x+h-3y-4z-12)}{h}\\right]}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{h\\to 0}(2x+h-3y-4z-12)}\\hfill \\\\ \\hfill & =\\hfill & {2x-3y-4z-12.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Then we find [latex]\\partial f\/\\partial y[\/latex] by holding [latex]x[\/latex] and [latex]z[\/latex] constant. Therefore, any term that does not include the variable [latex]y[\/latex] is constant, and its derivative is zero. We can apply the sum, difference, and power rules for functions of one variable:<\/p>\n<p>[latex]\\begin{alignat}{2} &\\frac{\\partial}{\\partial y}\\left[x^{2}-3xy+2y^{2}-4xz+5yz^{2}-12x+4y-3z\\right] \\\\    \\hspace{2cm}&=\\frac{\\partial}{\\partial y}\\left[x^{2}\\right]-\\frac{\\partial}{\\partial y}\\left[3xy\\right]+\\frac{\\partial}{\\partial y}\\left[2y^{2}\\right]-\\frac{\\partial}{\\partial y}\\left[4xz\\right]+\\frac{\\partial}{\\partial y}\\left[5yz^{2}\\right]-\\frac{\\partial}{\\partial y}\\left[12x\\right]+\\frac{\\partial}{\\partial y}\\left[4y\\right]-\\frac{\\partial}{\\partial y}\\left[3z\\right] &\\quad\\\\    &=0-3x+4y-0+5z^2-0+4-0\\\\    &=-3x+4y+5z^2+4.\\\\    \\end{alignat}[\/latex]<\/p>\n<p>To calculate\u00a0[latex]\\partial f\/\\partial z[\/latex], we hold [latex]x[\/latex] and [latex]y[\/latex] constant and apply the sum, difference, and power rules for functions of one variable:<\/p>\n<p>[latex]\\begin{alignat}{2} &\\frac{\\partial}{\\partial z}\\left[x^{2}-3xy+2y^{2}-4xz+5yz^{2}-12x+4y-3z\\right] \\\\    \\hspace{2cm}&=\\frac{\\partial}{\\partial z}\\left[x^{2}\\right]-\\frac{\\partial}{\\partial z}\\left[3xy\\right]+\\frac{\\partial}{\\partial z}\\left[2y^{2}\\right]-\\frac{\\partial}{\\partial z}\\left[4xz\\right]+\\frac{\\partial}{\\partial z}\\left[5yz^{2}\\right]-\\frac{\\partial}{\\partial z}\\left[12x\\right]+\\frac{\\partial}{\\partial z}\\left[4y\\right]-\\frac{\\partial}{\\partial y}\\left[3z\\right] &\\quad\\\\    &=0-0+0-4x+10yz-0+0-3\\\\    &=-4z+10yz-3.\\\\    \\end{alignat}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Calculate the three partial derivatives of the following functions.<\/p>\n<ol>\n<li>[latex]f\\,(x,\\ y,\\ z)=\\frac{x^{2}y-4xz+y^{2}}{x-3yz}[\/latex]<\/li>\n<li>[latex]g\\,(x,\\ y,\\ z)=\\sin{(x^{2}y-z)}+\\cos{(x^{2}-yz)}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1367703960124\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1367703960124\" class=\"hidden-answer\" style=\"display: none\">\n<div style=\"text-align: left;\">In each case, treat all variables as constants except the one whose partial derivative you are calculating.<\/div>\n<p>&nbsp;<\/p>\n<p style=\"padding-left: 30px;\">[latex]\\begin{array}{ccc}1. \\hspace{1cm} {\\frac{\\partial f}{\\partial x}} & =\\hfill & {\\frac{\\partial}{\\partial x}\\bigg[\\frac{x^{2}y-4xz+y^{2}}{x-3yz}\\bigg]}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{\\frac{\\partial}{\\partial x}(x^{2}y-4xz+y^{2})(x-3yz)-(x^{2}y-4xz+y^{2})\\frac{\\partial}{\\partial x}(x-3yz)}{(x-3yz)^{2}}}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{(2xy-4z)(x-3yz)-(x^{2}y-4xz+y^{2})(1)}{(x-3yz)^{2}}}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{2x^{2}y-6xy^{2}z-4xz+12yz^{2}-x^{2}y+4xz-y^{2}}{(x-3yz)^{2}}} \\hfill \\\\ \\hfill & =\\hfill & {\\frac{x^{2}y-6xy^{2}z-4xz+12yz^{2}+4xz-y^{2}}{(x-3yz)^{2}}}\\hfill \\\\ \\hfill {\\frac{\\partial f}{\\partial y}} & =\\hfill & {\\frac{\\partial}{\\partial y}\\bigg[\\frac{x^{2}y-4xz+y^{2}}{x-3yz}\\bigg]}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{\\frac{\\partial}{\\partial y}(x^{2}y-4xz+y^{2})(x-3yz)-(x^{2}y-4xz+y^{2})\\frac{\\partial}{\\partial y}(x-3yz)}{(x-3yz)^{2}}}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{(x^{2}+2y)(x-3yz)-(x^{2}y-4xz+y^{2})(-3z)}{(x-3yz)^{2}}}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{x^{3}-3x^{2}yz+2xy-6y^{2}z+3x^{2}yz-12xz^{2}+3y^{2}z}{(x-3yz)^{2}}}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{x^{3}+2xy-3y^{2}z-12xz^{2}}{(x-3yz)^{2}}}\\hfill \\\\ \\hfill {\\frac{\\partial f}{\\partial z}} & =\\hfill & {\\frac{\\partial}{\\partial z}\\bigg[\\frac{x^{2}y-4xz+y^{2}}{x-3yz}\\bigg]}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{\\frac{\\partial}{\\partial z}(x^{2}y-4xz+y^{2})(x-3yz)-(x^{2}y-4xz+y^{2})\\frac{\\partial}{\\partial z}(x-3yz)}{(x-3yz)^{2}}} \\hfill \\\\ \\hfill & =\\hfill & {\\frac{(-4x)(x-3yz)-(x^{2}y-4xz+y^{2})(-3y)}{(x-3yz)^{2}}} \\hfill \\\\ \\hfill & =\\hfill & {\\frac{-4x^{2}+12xyz+3x^{2}y^{2}-12xyz+3y^{3}}{(x-3yz)^{2}}}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{-4x^{2}+3x^{2}y^{2}+3y^{3}}{(x-3yz)^{2}}} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/p>\n<p style=\"padding-left: 30px;\">[latex]\\begin{array}{ccc}2. \\hspace{1cm} {\\frac{\\partial f}{\\partial x}} & =\\hfill & {\\frac{\\partial}{\\partial x}\\left[\\sin{(x^{2}y-z)}+\\cos{(x^{2}-yz)}\\right]}\\hfill \\\\ \\hfill & =\\hfill & {(\\cos{(x^{2}y-z)})\\frac{\\partial}{\\partial x}(x^{2}y-z)-(\\sin{(x^{2}-yz)})\\frac{\\partial}{\\partial x}(x^{2}-yz)}\\hfill \\\\ \\hfill & =\\hfill & {2xy\\cos{(x^{2}y-z)}-2x\\sin{(x^{2}-yz)}} \\hfill \\\\ \\hfill {\\frac{\\partial f}{\\partial y}} & =\\hfill & {\\frac{\\partial}{\\partial y}\\left[\\sin{(x^{2}y-z)}+\\cos{(x^{2}-yz)}\\right]}\\hfill \\\\ \\hfill & =\\hfill & {(\\cos{(x^{2}y-z)})\\frac{\\partial}{\\partial y}(x^{2}y-z)-(\\sin{(x^{2}-yz)})\\frac{\\partial}{\\partial y}(x^{2}-yz)}\\hfill \\\\ \\hfill & =\\hfill & {x^{2}\\cos{(x^{2}y-z)}+z\\sin{(x^{2}-yz)}} \\hfill \\\\ \\hfill {\\frac{\\partial f}{\\partial z}} & =\\hfill & {\\frac{\\partial}{\\partial z}\\left[\\sin{(x^{2}y-z)}+\\cos{(x^{2}-yz)}\\right]}\\hfill \\\\ \\hfill & =\\hfill & {(\\cos{(x^{2}y-z)})\\frac{\\partial}{\\partial z}(x^{2}y-z)-(\\sin{(x^{2}-yz)})\\frac{\\partial}{\\partial z}(x^{2}-yz)} \\hfill \\\\ \\hfill & =\\hfill & {-\\cos{(x^{2}y-z)}+y\\sin{(x^{2}-yz)}} \\hfill \\\\ \\hfill 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