{"id":1123,"date":"2025-10-15T14:31:04","date_gmt":"2025-10-15T14:31:04","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=1123"},"modified":"2025-10-27T21:05:39","modified_gmt":"2025-10-27T21:05:39","slug":"tangent-planes-and-linear-approximations-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/tangent-planes-and-linear-approximations-learn-it-3\/","title":{"raw":"Tangent Planes and Linear Approximations: Learn It 3","rendered":"Tangent Planes and Linear Approximations: Learn It 3"},"content":{"raw":"
Differentiability<\/h2>\r\n
For a function [latex]y=f(x)[\/latex] of one variable, we say the function is differentiable at [latex]x=a[\/latex] if [latex]f'(a)[\/latex] exists. When a function is differentiable at a point, the graph is \"smooth\" there\u2014no corners\u2014and a tangent line is well-defined.<\/p>\r\n
For functions of two variables, differentiability is also connected to smoothness. In this case, a surface is considered to be smooth at point [latex]P[\/latex] if a tangent plane to the surface exists at that point. If a function is differentiable at a point, then a tangent plane to the surface exists at that point.\u00a0 Recall that the tangent plane at [latex](x_0, y_0)[\/latex] is:<\/p>\r\n
For this tangent plane to exist, the partial derivatives must exist at [latex](x_0, y_0)[\/latex]. However, this isn't enough to guarantee smoothness, as was illustrated in Figure 3. The partial derivatives existed at the origin, but the function had a corner there.<\/p>\r\n\r\n\r\n
differentiable function at a point<\/h3>\r\nA function [latex]f(x, y)[\/latex] is differentiable<\/strong> at a point [latex]P\\ (x_0, y_0)[\/latex] if, for all points [latex](x, y)[\/latex] in a [latex]\\delta[\/latex] disk around [latex]P[\/latex], we can write:\r\n
[latex]f(x, y, z)=f(x_0, y_0)+f_x(x_0, y_0)(x-x_0)+f_y(x_0, y_0)(y-y_0)+E(x, y)[\/latex],<\/div>\r\n \r\n\r\nwhere the error term [latex]E[\/latex] satisfies:\r\n
[latex]\\displaystyle\\lim_{(x, y)\\to(x_0, y_0)}\\dfrac{E(x, y)}{\\sqrt{(x-x_0)^{2}+(y-y_0)^{2}}}=0.[\/latex]<\/div>\r\n<\/section>The error term [latex]E(x, y)[\/latex] represents how closely the tangent plane approximates the surface in a small neighborhood around point [latex]P[\/latex]. For [latex]f[\/latex] to be differentiable at [latex]P[\/latex], the function must be smooth\u2014the graph of [latex]f[\/latex] must stay close to the tangent plane for points near [latex]P[\/latex].\r\n\r\n\r\n
The function [latex]f(x, y)=\\begin{cases}\\dfrac{xy}{\\sqrt{x^{2}+y^{2}}} & (x, y) \\neq (0, 0) \\\\ 0 & (x, y)=(0, 0)\\end{cases}[\/latex] is not differentiable at the origin.<\/p>\r\n
We can see this by calculating the partial derivatives. This function appeared earlier in the section, where we showed that [latex]f_x(0, 0)=f_y(0, 0)=0[\/latex]. Substituting this information into our definition equation using [latex]x_0=0[\/latex] and [latex]y_0=0[\/latex], we get:<\/p>\r\n
Depending on the path taken toward the origin, this limit takes different values. Therefore, the limit does not exist and the function [latex]f[\/latex] is not differentiable at the origin as shown in the following figure.<\/p>\r\n\r\n[caption id=\"attachment_1240\" align=\"aligncenter\" width=\"579\"] Figure 1. This function [latex]f(x,y)[\/latex] is not differentiable at the origin.[\/caption]<\/section>Show that the function [latex]f(x, y)=2x^{2}-4y[\/latex] is differentiable at point [latex](2,-3)[\/latex].[reveal-answer q=\"fs-id1467793733125\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1467793733125\"]\r\n
First, we calculate [latex]f(x_0, y_0)[\/latex], [latex]f_x(x_0, y_0)[\/latex], and [latex]f_y(x_0, y_0)[\/latex] using [latex]x_0=2[\/latex] and [latex]y_0=-3[\/latex], then we use the equation for a Differentiable Function at a Point:<\/p>\r\n\r\n
[latex]\\begin{array}{ccc}\\hfill {f(2,-3)} & =\\hfill & {2(2)^{2}-4(-3)=8+12=20}\\hfill \\\\ \\hfill {f_x(2,-3)} & =\\hfill & {4(2)=8}\\hfill \\\\ \\hfill {f_y(2,-3)} & =\\hfill & {-4.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n \r\n\r\nTherefore [latex]m_1=8[\/latex] and [latex]m_2 =-4[\/latex], and the equation for a Differentiable Function at a Point becomes\r\n
[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\lim_{(x, y)\\to (x_0, y_0)}\\frac{E(x, y)}{\\sqrt{(x-x_0)^{2}+(y-y_0)^{2}}}} & =\\hfill & {\\displaystyle\\lim_{(x, y)\\to (2,-3)}\\frac{2x^{2}-8x+8}{\\sqrt{(x-2)^{2}+(y+3)^{2}}}}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{(x, y)\\to (2,-3)}\\frac{2(x^{2}-4x+4)}{\\sqrt{(x-2)^{2}+(y+3)^{2}}}}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{(x, y)\\to (2,-3)}\\frac{2(x-2)^{2}}{\\sqrt{(x-2)^{2}+(y+3)^{2}}}}\\hfill \\\\ \\hfill & \\leq\\hfill & {\\displaystyle\\lim_{(x, y)\\to (2,-3)}\\frac{2((x-2)^{2}+(y+3)^{2})}{\\sqrt{(x-2)^{2}+(y+3)^{2}}}} \\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{(x, y)\\to (2,-3)}2\\sqrt{(x-2)^{2}+(y+3)^{2}}}\\hfill \\\\ \\hfill & =\\hfill & {0.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n \r\n\r\nSince [latex]E(x, y)\\geq 0[\/latex] for any value of [latex]x[\/latex] or [latex]y[\/latex], the original limit must be equal to zero. Therefore, [latex]f(x, y)=2x^{2}-4y[\/latex] is differentiable at point [latex](2,-3)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n
The Connection Between Differentiability and Continuity<\/h3>\r\nDifferentiability and continuity for functions of two or more variables are connected, just as they are for functions of one variable.\r\n\r\n\r\n
differentiability implies continuity<\/h3>\r\nLet [latex]z=f(x, y)[\/latex] be a function of two variables with [latex](x_0, y_0)[\/latex] in the domain of [latex]f[\/latex]. If [latex]f(x, y)[\/latex] is differentiable at [latex](x_0, y_0)[\/latex], then [latex]f(x, y)[\/latex] is continuous at [latex](x_0, y_0)[\/latex].\r\n\r\n<\/section>\r\n
This theorem shows that if a function is differentiable at a point, then it must be continuous there. However, the reverse isn't true\u2014a function can be continuous at a point without being differentiable there. For example,<\/p>\r\n
is continuous at the origin, but not differentiable there. This is similar to what happens in single-variable calculus.<\/p>\r\n
The next theorem gives us a practical way to check for differentiability. It says that if the function and its partial derivatives are all continuous at a point, then the function is differentiable there.<\/p>\r\n\r\n\r\n
continuity of first partials implies differentiability<\/h3>\r\nLet [latex]z=f(x, y)[\/latex] be a function of two variables with [latex](x_0, y_0)[\/latex] in the domain of [latex]f[\/latex]. If [latex]f(x, y)[\/latex], [latex]f_x(x, y)[\/latex], and [latex]f_y(x, y)[\/latex] all exist in a neighborhood of [latex](x_0, y_0)[\/latex] and are continuous at [latex](x_0, y_0)[\/latex], then [latex]f(x, y)[\/latex] is differentiable there.\r\n\r\n<\/section>\r\n
Let's revisit the function we showed wasn't differentiable at the origin:<\/p>\r\n
[latex]\\dfrac{\\partial f}{\\partial x}=\\dfrac{y^{3}}{(x^{2}+y^{2})^{3\/2}}[\/latex] and [latex]\\dfrac{\\partial f}{\\partial y}=\\dfrac{x^{3}}{(x^{2}+y^{2})^{3\/2}}[\/latex]<\/p>\r\n
The contrapositive of the previous theorem states that if a function is not differentiable, then at least one of the hypotheses must fail. Let's check whether [latex]f_x[\/latex] is continuous at [latex](0,0)[\/latex]. For continuity, we need [latex]\\displaystyle\\lim_{(x,y)\\to(0,0)}{f_x(x,y) = f_x(0,0)}[\/latex]:<\/p>\r\n
[latex]\\begin{array}{ccc}\\hfill\\displaystyle\\lim_{(x,y)\\to(0,0)}{\\frac{y^3}{(x^2+y^2)^{3\/2}}}&=\\displaystyle\\lim_{y\\to0}{\\frac{y^3}{((ky)^2+y^2)^{3\/2}}}\\\\&=\\displaystyle\\lim_{y\\to0}{\\frac{y^3}{(k^2y^2+y^2)^{3\/2}}}&\\quad\\\\&=\\displaystyle\\lim_{y\\to0}{\\frac{y^3}{|y^3|(k^2+1)^{3\/2}}}\\\\&=\\frac1{(k^2+1)^{3\/2}}\\displaystyle\\lim_{y\\to0}{\\frac{|y|}{y}}\\\\\\end{array}[\/latex]<\/p>\r\nIf [latex]y>0[\/latex], this expression equals [latex]1\/(k^2+1)^{3\/2}[\/latex]; if [latex]y<0[\/latex], it equals [latex]-\\left(1\/(k^2+1)^{3\/2}\\right)[\/latex]. In either case, the value depends on [latex]k[\/latex], so the limit fails to exist. This confirms that [latex]f_x[\/latex] is not continuous at the origin, which is why the function isn't differentiable there.","rendered":"
Differentiability<\/h2>\n
For a function [latex]y=f(x)[\/latex] of one variable, we say the function is differentiable at [latex]x=a[\/latex] if [latex]f'(a)[\/latex] exists. When a function is differentiable at a point, the graph is “smooth” there\u2014no corners\u2014and a tangent line is well-defined.<\/p>\n
For functions of two variables, differentiability is also connected to smoothness. In this case, a surface is considered to be smooth at point [latex]P[\/latex] if a tangent plane to the surface exists at that point. If a function is differentiable at a point, then a tangent plane to the surface exists at that point.\u00a0 Recall that the tangent plane at [latex](x_0, y_0)[\/latex] is:<\/p>\n
For this tangent plane to exist, the partial derivatives must exist at [latex](x_0, y_0)[\/latex]. However, this isn’t enough to guarantee smoothness, as was illustrated in Figure 3. The partial derivatives existed at the origin, but the function had a corner there.<\/p>\n\n
differentiable function at a point<\/h3>\n
A function [latex]f(x, y)[\/latex] is differentiable<\/strong> at a point [latex]P\\ (x_0, y_0)[\/latex] if, for all points [latex](x, y)[\/latex] in a [latex]\\delta[\/latex] disk around [latex]P[\/latex], we can write:<\/p>\n
[latex]f(x, y, z)=f(x_0, y_0)+f_x(x_0, y_0)(x-x_0)+f_y(x_0, y_0)(y-y_0)+E(x, y)[\/latex],<\/div>\n
<\/p>\n
where the error term [latex]E[\/latex] satisfies:<\/p>\n
The error term [latex]E(x, y)[\/latex] represents how closely the tangent plane approximates the surface in a small neighborhood around point [latex]P[\/latex]. For [latex]f[\/latex] to be differentiable at [latex]P[\/latex], the function must be smooth\u2014the graph of [latex]f[\/latex] must stay close to the tangent plane for points near [latex]P[\/latex].<\/p>\n\n
The function [latex]f(x, y)=\\begin{cases}\\dfrac{xy}{\\sqrt{x^{2}+y^{2}}} & (x, y) \\neq (0, 0) \\\\ 0 & (x, y)=(0, 0)\\end{cases}[\/latex] is not differentiable at the origin.<\/p>\n
We can see this by calculating the partial derivatives. This function appeared earlier in the section, where we showed that [latex]f_x(0, 0)=f_y(0, 0)=0[\/latex]. Substituting this information into our definition equation using [latex]x_0=0[\/latex] and [latex]y_0=0[\/latex], we get:<\/p>\n
Depending on the path taken toward the origin, this limit takes different values. Therefore, the limit does not exist and the function [latex]f[\/latex] is not differentiable at the origin as shown in the following figure.<\/p>\nFigure 1. This function [latex]f(x,y)[\/latex] is not differentiable at the origin.<\/figcaption><\/figure>\n<\/section>\nShow that the function [latex]f(x, y)=2x^{2}-4y[\/latex] is differentiable at point [latex](2,-3)[\/latex].<\/p>\n
![\"A](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/22220707\/4-4-6.jpeg\")
Figure 1. This function [latex]f(x,y)[\/latex] is not differentiable at the origin.[\/caption]<\/section>