{"id":599,"date":"2025-10-09T15:24:35","date_gmt":"2025-10-09T15:24:35","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=599"},"modified":"2026-05-26T14:59:37","modified_gmt":"2026-05-26T14:59:37","slug":"introduction-to-partial-derivatives-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/introduction-to-partial-derivatives-learn-it-1\/","title":{"raw":"Introduction to Partial Derivatives: Learn It 1","rendered":"Introduction to Partial Derivatives: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Calculate the partial derivatives of a function of two variables<\/li>\r\n \t<li>Calculate the partial derivatives of a function of more than two variables<\/li>\r\n \t<li>Find higher-order partial derivatives of a function of two variables<\/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]325842[\/ohm_question]<\/section>\r\n<h3 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">Geometric Interpretation<\/h3>\r\n<p class=\"whitespace-normal break-words\">How should we interpret partial derivatives geometrically? Recall that the graph of a function of two variables is a surface in [latex]\\mathbb{R}^{3}[\/latex]. If we remove the limit from the definition of the partial derivative with respect to [latex]x[\/latex], we're left with the difference quotient:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\dfrac{f(x+h,y)-f(x,y)}{h}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">This looks just like the difference quotient for a single-variable derivative, except [latex]y[\/latex] is along for the ride as a constant. Figure 1 shows a surface described by [latex]z=f(x,y)[\/latex].<\/p>\r\n\r\n[caption id=\"attachment_981\" align=\"aligncenter\" width=\"468\"]<img class=\"size-full wp-image-981\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/28225847\/4-3-1.jpeg\" alt=\"A complicated curve in xyz space with a secant line through the points (x, y, f(x, y)) and (x + h, y, f(x + h, y)).\" width=\"468\" height=\"395\" \/> Figure 1. Secant line passing through the points [latex]\\small{(x,y,f(x,y))}[\/latex] and [latex]\\small{(x+h,y,f(x+h,y))}[\/latex].[\/caption]\r\n<p class=\"whitespace-normal break-words\">In Figure 1, [latex]h[\/latex] is positive. When we graph the points [latex]f(x, y)[\/latex] and [latex]f(x+h, y)[\/latex], the slope of the secant line connecting them is:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\dfrac{f(x+h,y)-f(x,y)}{h}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">This secant line runs parallel to the [latex]x[\/latex]-axis. The slope represents an average rate of change of [latex]f[\/latex] as we move parallel to the [latex]x[\/latex]-axis. As [latex]h[\/latex] approaches zero, the secant line's slope approaches the tangent line's slope.<\/p>\r\n<p class=\"whitespace-normal break-words\">If we instead change [latex]y[\/latex] by an increment [latex]h[\/latex] (while holding [latex]x[\/latex] fixed), the secant line\u2014and resulting tangent line\u2014would be parallel to the [latex]y[\/latex]-axis.<\/p>\r\n<p class=\"whitespace-normal break-words\">Therefore:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{\\partial f}{\\partial x}[\/latex] gives the slope of the tangent line to the surface parallel to the [latex]x[\/latex]-axis at the point [latex](x, y, f(x, y))[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{\\partial f}{\\partial y}[\/latex] gives the slope of the tangent line to the surface parallel to the [latex]y[\/latex]-axis at the point [latex](x, y, f(x, y))[\/latex]<\/li>\r\n<\/ul>\r\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">\r\n<p class=\"whitespace-normal break-words\">Think of partial derivatives as slopes in specific directions:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{\\partial f}{\\partial x}[\/latex]: slope when moving parallel to the [latex]x[\/latex]-axis<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{\\partial f}{\\partial y}[\/latex]: slope when moving parallel to the [latex]y[\/latex]-axis<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-pre-wrap break-words\">What about slopes in other directions? We'll need directional derivatives for that, which we'll discuss later.<\/p>\r\n\r\n<\/section>\r\n<div>\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0 !gap-3.5\">\r\n<p class=\"whitespace-normal break-words\">You can also estimate partial derivatives using contour maps.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Use a contour map to estimate [latex]\\partial g\/\\partial x[\/latex] at the point [latex](\\sqrt{5}, 0)[\/latex] for the function\r\n<div style=\"text-align: center;\">[latex]g(x, y)=\\sqrt{9-x^{2}-y^{2}}[\/latex].<\/div>\r\n<div>[reveal-answer q=\"fs-id1167793943114\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793943114\"]\r\n<p style=\"text-align: left;\">The following graph represents a contour map for the function [latex]g(x, y)=\\sqrt{9-x^{2}-y^{2}}[\/latex].<\/p>\r\n\r\n<div>[caption id=\"attachment_983\" align=\"aligncenter\" width=\"344\"]<img class=\"size-full wp-image-983\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/28230034\/4-3-2.jpeg\" alt=\"A series of concentric circles with the center the origin. The first is marked c = 0 and has radius 3; the second is marked c = 1 and has radius slightly less than 3; and the third is marked c = 2 and has radius slightly more than 2. The graph is marked with the equation g(x, y) = the square root of the quantity (9 \u2013 x2 \u2013 y2).\" width=\"344\" height=\"347\" \/> Figure 2. Contour map for the function [latex]\\small{g(x,y)=\\sqrt{9-x^{2}-y^{2}}}[\/latex], using [latex]\\small{c=0,1,2}[\/latex], and [latex]\\small{3}[\/latex] ([latex]\\small{c=3}[\/latex] corresponds to the origin).[\/caption]<\/div>\r\nThe inner circle on the contour map corresponds to [latex]c=2[\/latex] and the next circle out corresponds to [latex]c=1[\/latex]. The first circle is given by the equation [latex]2=\\sqrt{9-x^{2}-y^{2}}[\/latex]; the second circle is given by the equation [latex]1=\\sqrt{9-x^{2}-y^{2}}[\/latex]. The first equation simplifies to [latex]x^{2}+y^{2}=5[\/latex] and the second equation simplifies to [latex]x^{2}+y^{2}=8[\/latex]. The [latex]x[\/latex]-intercept of the first circle is [latex](\\sqrt{5}, 0)[\/latex] and the [latex]x[\/latex]-intercept of the second circle is [latex](2\\sqrt{2}, 0)[\/latex]. We can estimate the value of [latex]\\partial g\/\\partial x[\/latex] evaluated at the point [latex](\\sqrt{5}, 0)[\/latex] using the slope formula:\r\n<div style=\"text-align: center;\">[latex]\\large{\\frac{\\partial g}{\\partial x}\\bigg|_{(x, y)=(\\sqrt{5}, 0)}\\approx\\frac{g(\\sqrt{5}, 0)-g(2\\sqrt{2}, 0)}{\\sqrt{5}-2\\sqrt{2}}=\\frac{2-1}{\\sqrt{5}-2\\sqrt{2}}=\\frac{1}{\\sqrt{5}-2\\sqrt{2}}\\approx -1.688.}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nTo calculate the exact value of [latex]\\partial g\/\\partial x[\/latex] evaluated at the point [latex](\\sqrt{5}, 0)[\/latex], we start by finding [latex]\\partial g\/\\partial x[\/latex] using the chain rule. First, we rewrite the function as [latex]g(x, y)=\\sqrt{9-x^{2}-y^{2}}=(9-x^{2}-y^{2})^{\\frac{1}{2}}[\/latex] and then differentiate with respect to [latex]x[\/latex] while holding [latex]y[\/latex] constant:\r\n<div style=\"text-align: center;\">[latex]\\large{\\frac{\\partial g}{\\partial x}=\\frac{1}{2}(9-x^{2}-y^{2})^{-\\frac{1}{2}}(-2x)=-\\frac{x}{\\sqrt{9-x^{2}-y^{2}}}.}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nNext, we evaluate this expression using [latex]x=\\sqrt{5}[\/latex] and [latex]y=0[\/latex]:\r\n<div style=\"text-align: center;\">[latex]\\large{\\frac{\\partial g}{\\partial x}\\bigg|_{(x, y)=(\\sqrt{5}, 0)}=-\\frac{\\sqrt{5}}{\\sqrt{9-(\\sqrt{5})^{2}-(0)^{2}}}=-\\frac{\\sqrt{5}}{\\sqrt{4}}=-\\frac{\\sqrt{5}}{2}\\approx -1.118.}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nThe estimate for the partial derivative corresponds to the slope of the secant line passing through the points [latex](\\sqrt{5}, 0, g(\\sqrt{5}, 0))[\/latex] and [latex](2\\sqrt{2}, 0, g(2\\sqrt{2}, 0))[\/latex]. It represents an approximation to the slope of the tangent line to the surface through the point [latex](\\sqrt{5}, 0, g(\\sqrt{5}, 0))[\/latex], which is parallel to the [latex]x[\/latex]-axis.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Calculate the partial derivatives of a function of two variables<\/li>\n<li>Calculate the partial derivatives of a function of more than two variables<\/li>\n<li>Find higher-order partial derivatives of a function of two variables<\/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=\"ohm325842\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=325842&theme=lumen&iframe_resize_id=ohm325842&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<h3 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">Geometric Interpretation<\/h3>\n<p class=\"whitespace-normal break-words\">How should we interpret partial derivatives geometrically? Recall that the graph of a function of two variables is a surface in [latex]\\mathbb{R}^{3}[\/latex]. If we remove the limit from the definition of the partial derivative with respect to [latex]x[\/latex], we&#8217;re left with the difference quotient:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\dfrac{f(x+h,y)-f(x,y)}{h}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">This looks just like the difference quotient for a single-variable derivative, except [latex]y[\/latex] is along for the ride as a constant. Figure 1 shows a surface described by [latex]z=f(x,y)[\/latex].<\/p>\n<figure id=\"attachment_981\" aria-describedby=\"caption-attachment-981\" style=\"width: 468px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-981\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/28225847\/4-3-1.jpeg\" alt=\"A complicated curve in xyz space with a secant line through the points (x, y, f(x, y)) and (x + h, y, f(x + h, y)).\" width=\"468\" height=\"395\" \/><figcaption id=\"caption-attachment-981\" class=\"wp-caption-text\">Figure 1. Secant line passing through the points [latex]\\small{(x,y,f(x,y))}[\/latex] and [latex]\\small{(x+h,y,f(x+h,y))}[\/latex].<\/figcaption><\/figure>\n<p class=\"whitespace-normal break-words\">In Figure 1, [latex]h[\/latex] is positive. When we graph the points [latex]f(x, y)[\/latex] and [latex]f(x+h, y)[\/latex], the slope of the secant line connecting them is:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\dfrac{f(x+h,y)-f(x,y)}{h}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">This secant line runs parallel to the [latex]x[\/latex]-axis. The slope represents an average rate of change of [latex]f[\/latex] as we move parallel to the [latex]x[\/latex]-axis. As [latex]h[\/latex] approaches zero, the secant line&#8217;s slope approaches the tangent line&#8217;s slope.<\/p>\n<p class=\"whitespace-normal break-words\">If we instead change [latex]y[\/latex] by an increment [latex]h[\/latex] (while holding [latex]x[\/latex] fixed), the secant line\u2014and resulting tangent line\u2014would be parallel to the [latex]y[\/latex]-axis.<\/p>\n<p class=\"whitespace-normal break-words\">Therefore:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{\\partial f}{\\partial x}[\/latex] gives the slope of the tangent line to the surface parallel to the [latex]x[\/latex]-axis at the point [latex](x, y, f(x, y))[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{\\partial f}{\\partial y}[\/latex] gives the slope of the tangent line to the surface parallel to the [latex]y[\/latex]-axis at the point [latex](x, y, f(x, y))[\/latex]<\/li>\n<\/ul>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">\n<p class=\"whitespace-normal break-words\">Think of partial derivatives as slopes in specific directions:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{\\partial f}{\\partial x}[\/latex]: slope when moving parallel to the [latex]x[\/latex]-axis<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{\\partial f}{\\partial y}[\/latex]: slope when moving parallel to the [latex]y[\/latex]-axis<\/li>\n<\/ul>\n<p class=\"whitespace-pre-wrap break-words\">What about slopes in other directions? We&#8217;ll need directional derivatives for that, which we&#8217;ll discuss later.<\/p>\n<\/section>\n<div>\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0 !gap-3.5\">\n<p class=\"whitespace-normal break-words\">You can also estimate partial derivatives using contour maps.<\/p>\n<\/div>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Use a contour map to estimate [latex]\\partial g\/\\partial x[\/latex] at the point [latex](\\sqrt{5}, 0)[\/latex] for the function<\/p>\n<div style=\"text-align: center;\">[latex]g(x, y)=\\sqrt{9-x^{2}-y^{2}}[\/latex].<\/div>\n<div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167793943114\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167793943114\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: left;\">The following graph represents a contour map for the function [latex]g(x, y)=\\sqrt{9-x^{2}-y^{2}}[\/latex].<\/p>\n<div>\n<figure id=\"attachment_983\" aria-describedby=\"caption-attachment-983\" style=\"width: 344px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-983\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/28230034\/4-3-2.jpeg\" alt=\"A series of concentric circles with the center the origin. The first is marked c = 0 and has radius 3; the second is marked c = 1 and has radius slightly less than 3; and the third is marked c = 2 and has radius slightly more than 2. The graph is marked with the equation g(x, y) = the square root of the quantity (9 \u2013 x2 \u2013 y2).\" width=\"344\" height=\"347\" \/><figcaption id=\"caption-attachment-983\" class=\"wp-caption-text\">Figure 2. Contour map for the function [latex]\\small{g(x,y)=\\sqrt{9-x^{2}-y^{2}}}[\/latex], using [latex]\\small{c=0,1,2}[\/latex], and [latex]\\small{3}[\/latex] ([latex]\\small{c=3}[\/latex] corresponds to the origin).<\/figcaption><\/figure>\n<\/div>\n<p>The inner circle on the contour map corresponds to [latex]c=2[\/latex] and the next circle out corresponds to [latex]c=1[\/latex]. The first circle is given by the equation [latex]2=\\sqrt{9-x^{2}-y^{2}}[\/latex]; the second circle is given by the equation [latex]1=\\sqrt{9-x^{2}-y^{2}}[\/latex]. The first equation simplifies to [latex]x^{2}+y^{2}=5[\/latex] and the second equation simplifies to [latex]x^{2}+y^{2}=8[\/latex]. The [latex]x[\/latex]-intercept of the first circle is [latex](\\sqrt{5}, 0)[\/latex] and the [latex]x[\/latex]-intercept of the second circle is [latex](2\\sqrt{2}, 0)[\/latex]. We can estimate the value of [latex]\\partial g\/\\partial x[\/latex] evaluated at the point [latex](\\sqrt{5}, 0)[\/latex] using the slope formula:<\/p>\n<div style=\"text-align: center;\">[latex]\\large{\\frac{\\partial g}{\\partial x}\\bigg|_{(x, y)=(\\sqrt{5}, 0)}\\approx\\frac{g(\\sqrt{5}, 0)-g(2\\sqrt{2}, 0)}{\\sqrt{5}-2\\sqrt{2}}=\\frac{2-1}{\\sqrt{5}-2\\sqrt{2}}=\\frac{1}{\\sqrt{5}-2\\sqrt{2}}\\approx -1.688.}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>To calculate the exact value of [latex]\\partial g\/\\partial x[\/latex] evaluated at the point [latex](\\sqrt{5}, 0)[\/latex], we start by finding [latex]\\partial g\/\\partial x[\/latex] using the chain rule. First, we rewrite the function as [latex]g(x, y)=\\sqrt{9-x^{2}-y^{2}}=(9-x^{2}-y^{2})^{\\frac{1}{2}}[\/latex] and then differentiate with respect to [latex]x[\/latex] while holding [latex]y[\/latex] constant:<\/p>\n<div style=\"text-align: center;\">[latex]\\large{\\frac{\\partial g}{\\partial x}=\\frac{1}{2}(9-x^{2}-y^{2})^{-\\frac{1}{2}}(-2x)=-\\frac{x}{\\sqrt{9-x^{2}-y^{2}}}.}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Next, we evaluate this expression using [latex]x=\\sqrt{5}[\/latex] and [latex]y=0[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]\\large{\\frac{\\partial g}{\\partial x}\\bigg|_{(x, y)=(\\sqrt{5}, 0)}=-\\frac{\\sqrt{5}}{\\sqrt{9-(\\sqrt{5})^{2}-(0)^{2}}}=-\\frac{\\sqrt{5}}{\\sqrt{4}}=-\\frac{\\sqrt{5}}{2}\\approx -1.118.}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>The estimate for the partial derivative corresponds to the slope of the secant line passing through the points [latex](\\sqrt{5}, 0, g(\\sqrt{5}, 0))[\/latex] and [latex](2\\sqrt{2}, 0, g(2\\sqrt{2}, 0))[\/latex]. It represents an approximation to the slope of the tangent line to the surface through the point [latex](\\sqrt{5}, 0, g(\\sqrt{5}, 0))[\/latex], which is parallel to the [latex]x[\/latex]-axis.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":17,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-599","chapter","type-chapter","status-publish","hentry"],"part":426,"module-header":"learn_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/599","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":10,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/599\/revisions"}],"predecessor-version":[{"id":2885,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/599\/revisions\/2885"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/426"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/599\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=599"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=599"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=599"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=599"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}