2.16 Implicit Differentiation

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We learn to compute the derivative of an implicit function.

Implicit Differentiation

Implicit Differentiation is used to find $\frac {dy}{dx}$ in situations where $y$ is not written as an explicit function of $x$ . Some examples of equations where implicit differentiation is necessary are:

$x^2 + y^2 = 1, \text { the unit circle}$ $4x^2 + 9y^2 = 36, \text { an ellipse}$ $x^3 + y^3 =6xy, \text { the Folium of Descartes}$ $x^2(x^2 + y^2) =y^2, \text { the kappa curve and}$ $(x^2 + y^2)^2 =50xy, \text { a lemniscate}.$

To compute $\frac {dy}{dx}$ in these situations, we make the assumption that $y$ is an unspecified function of $x$ and in most cases, we employ the chain rule with $y$ as the inside function.

Warm-up Examples of Implicit Differentiation

example 1 Find $\frac {d}{dx} y^2$ where $y$ is an unspecified function of $x$ .
We use the chain rule, $\displaystyle {[f(g(x))]' = f'(g(x))g'(x)}$ with $g(x) = y$ and $f(x) = x^2$ . Then $g'(x) = \frac {dy}{dx} \text { and}$ $f'(x) = 2x, \text { so } f'(g(x)) = 2g(x) = 2y$ and we can conclude that $\frac {d}{dx} y^2 = 2y \frac {dy}{dx}.$

Here is a video of Warm-up 1

(problem 1) Compute $\displaystyle {\frac {d}{dx} \left (y^3\right )=}$

$3y^2$ $3y^2 \frac {d}{dx}$ $3y \frac {dy}{dx}$ $3y^2 \frac {dy}{dx}$

Treat $y$ as an unspecified function of $x$ , i.e., $y = g(x)$ .

Use the Chain Rule with $y$ as the inside function.

example 2 Find $\frac {d}{dx} \sin (y)$ where $y$ is an unspecified function of $x$ .
We use the chain rule, $\displaystyle {[f(g(x))]' = f'(g(x))g'(x)}$ with $g(x) = y$ and $f(x) = \sin (x)$ . Then $g'(x) = \frac {dy}{dx} \text { and}$ $f'(x) = \cos (x), \text { so } f'(g(x)) = \cos (g(x)) = \cos (y)$ and we can conclude that $\frac {d}{dx} \sin (y) = \cos (y) \frac {dy}{dx}.$

Here is a video of Warm-up 2

(problem 2a) Compute $\displaystyle {\frac {d}{dx} \left [\cos (y)\right ]=}$

$\sin (y)$ $-\sin (y) \frac {d}{dx}$ $-\sin (y) \frac {dy}{dx}$ $\sin (y) \frac {dy}{dx}$

Consider $y$ to be a function of $x$ .

Use the chain rule with $y$ as the inside function.

(problem 2b) $\displaystyle {\frac {d}{dx} \left [\sec (y)\right ]=}$

$\sec (y)\tan (y)$ $\sec ^2(y)\frac {d}{dx}$ $\tan ^2(y) \frac {dy}{dx}$ $\sec (y)\tan (y) \frac {dy}{dx}$

Consider $y$ to be a function of $x$ .

The derivative of $\sec (x)$ with respect to x is $\sec (x)\tan (x)$ .

Use the chain rule with $y$ as the inside function.

(problem 2c) $\displaystyle {\frac {d}{dx} \left (e^y\right )=}$

$ye^{y-1}$ $e^y\frac {d}{dx}$ $e^y \frac {dy}{dx}$ $ye^{y-1} \frac {dy}{dx}$

(problem 2d) $\displaystyle {\frac {d}{dx} \left (3^y\right )=}$

$3^y \ln (3)$ $3^y \ln (y)\frac {d}{dx}$ $3^y \ln (3) \frac {dy}{dx}$ $3^y \ln (y) \frac {dy}{dx}$

$\frac {d}{dx}\left (a^x\right )=a^x\ln (a)$

(problem 2e) $\displaystyle {\frac {d}{dx} \left [\ln (y)\right ]=}$

$\frac {1}{y}$ $\frac {1}{y}\frac {d}{dx}$ $\frac {1}{x}(y) \frac {dy}{dx}$ $\frac {1}{y} \frac {dy}{dx}$

(problem 2f) $\displaystyle {\frac {d}{dx} \left [\log (y)\right ]=}$

$\frac {1}{y}$ $\frac {1}{y\ln (10)}\frac {d}{dx}$ $\frac {1}{x}y\ln (10) \frac {dy}{dx}$ $\frac {1}{y\ln (10)} \frac {dy}{dx}$

$\frac {d}{dx}\left [\log _a(x)\right ] = \frac {1}{x\ln (a)}$

(problem 2g) $\displaystyle {\frac {d}{dx} \left [\tan ^{-1}(y)\right ]=}$

$\frac {1}{1+y^2}$ $\frac {1}{1-y^2}\frac {d}{dx}$ $\frac {1}{1+y^2} \frac {dy}{dx}$ $\frac {1}{1-y^2} \frac {dy}{dx}$

$\frac {d}{dx}\left [\\tan^{-1}(x)\right ] = \frac {1}{1+x^2}$

example 3 Find $\frac {d}{dx} \left (\frac {1}{y}\right )$ where $y$ is an unspecified function of $x$ .
We use the chain rule, $\displaystyle {[f(g(x))]' = f'(g(x))g'(x)}$ with $g(x) = y$ and $f(x) = \frac {1}{x}$ . Then $g'(x) = \frac {dy}{dx} \text { and}$ $f'(x) = -\frac {1}{x^2}, \text { so } f'(g(x)) = -\frac {1}{g^2(x)} = -\frac {1}{y^2}$ and we can conclude that $\frac {d}{dx} \Big (\frac {1}{y}\Big ) = -\frac {1}{y^2} \frac {dy}{dx}.$

Here is a video of Warm-up 3

(problem 3) $\displaystyle {\frac {d}{dx} \left (\frac {1}{y^2}\right )=}$

$\frac {1}{2y}$ $\frac {1}{2y}\frac {dy}{dx}$ $-\frac {2}{y^3}\frac {dy}{dx}$ $\frac {2}{y^3}\frac {dy}{dx}$

$\frac {1}{y^2} = y^{-2}$

example 4 Find $\frac {d}{dx} (6xy)$ where $y$ is an unspecified function of $x$ .
We use the product rule, $\displaystyle {[f(x) + g(x)]' = f(x)g'(x) + g(x)f'(x)}$ with $f(x) = 6x$ and $g(x) = y$ . Then $f'(x) = 6 \text { and } g'(x) = \frac {dy}{dx}.$ We conclude that $\frac {d}{dx} (6xy) = 6x \frac {dy}{dx} + 6y.$

Here is a video of Warm-up 4

(problem 4) $\displaystyle {\frac {d}{dx} \left (3x^2y\right )=}$

$6x$ $6x + \frac {dy}{dx}$ $3x^2 + 6xy$ $3x^2 \frac {dy}{dx} + 6xy$

Use the product rule

example 5 Find $\frac {d}{dx} (x^2y^3)$ where $y$ is an unspecified function of $x$ .
We use the product rule,

$[f(x) + g(x)]' = f(x)g'(x) + g(x)f'(x)$ with $f(x) = x^2$ and $g(x) = y^3$ . Then $f'(x) = 2x$ and by the chain rule, $g'(x) = 3y^2\frac {dy}{dx}.$ We conclude that $\frac {d}{dx} (x^2y^3) = 3x^2y^2\frac {dy}{dx} + 2xy^3.$

Here is a video of Warm-up 5

(problem 5a) $\displaystyle {\frac {d}{dx} \left (x^3y^4\right )=}$

$12x^2y^3$ $12x^2y^3\frac {dy}{dx}$ $3x^2y^4 + 4x^3y^3$ $3x^2 y^4 + 4x^3y^3 \frac {dy}{dx}$

Use the product rule and the chain rule

(problem 5b) $\displaystyle {\frac {d}{dx} \left (x^2e^y\right )=}$

$2xe^y$ $2xe^y\frac {dy}{dx}$ $2xe^y + x^2e^y$ $2xe^y + x^2e^y\frac {dy}{dx}$

Use the product rule and the chain rule

We are now ready to find $\frac {dy}{dx}$ in situations where the formula describing a curve in the $x,y$ -plane is not given in the form $y = f(x)$ .

Examples of Implicit Differentiation

example 6 Find $\displaystyle {\frac {dy}{dx}}$ if $x^2 + y^2 = 1$ .
We assume that $y$ is a function of $x$ and we differentiate both sides with respect to $x$ : $\frac {d}{dx}(x^2 + y^2) = \frac {d}{dx} (1)$ $\frac {d}{dx}(x^2) + \frac {d}{dx} (y^2) = 0$ $2x + 2y\frac {dy}{dx} = 0.$ We can now solve for $\displaystyle {\frac {dy}{dx}}$ algebraically: $2y\frac {dy}{dx} = -2x$ $\frac {dy}{dx} = -\frac {2x}{2y}= -\frac {x}{y}.$

Here is a video of Example 1

(problem 6a) Use the result of the example above to find the equation of the tangent line to the unit circle $x^2 + y^2 = 1$ at the point $(-\tfrac 35,\tfrac 45).$

$y-y_0 = m(x-x_0)$

$m = \frac {dy}{dx}\bigg |_{(-\frac 35, \frac 45)}$

The equation of the tangent line is $y = \answer {\frac 34 x+\frac 54}.$

(problem 6b) Find $\frac {dy}{dx}$ if $x^2 + y^2 = 25$

Consider $y$ to be a function of $x$ .

Differentiate both sides with respect to $x$ .

Use the chain rule on $y^2$ with $y$ as the inside function.

The derivative of the inside is $\frac {dy}{dx}$

$2x + 2y\frac {dy}{dx} = 0$

Solve for $\frac {dy}{dx}$ by moving the terms with $\frac {dy}{dx}$ to one side and the rest of the terms to the other side.

If $x^2 + y^2 = 25$ then $\frac {dy}{dx}$ is $\answer {-\frac {x}{y}}$

example 7 Find $\frac {dy}{dx}$ if $4x^2 + 9y^2 = 36$ .
We assume that $y$ is a function of $x$ and we differentiate both sides with respect to $x$ : $\frac {d}{dx}(4x^2 + 9y^2) = \frac {d}{dx} (36)$ $\frac {d}{dx}(4x^2) + \frac {d}{dx} (9y^2) = 0$ $8x + 18y\frac {dy}{dx} = 0.$ We can now solve for $\displaystyle {\frac {dy}{dx}}$ algebraically: $18y\frac {dy}{dx} = -8x$ $\frac {dy}{dx} = -\frac {8x}{18y}= -\frac {4x}{9y}.$

Here is a video of Example 2

(problem 7a) Find $\frac {dy}{dx}$ if $3x^2 + 5y^2 = 11$

Consider $y$ to be a function of $x$ .

Differentiate both sides with respect to $x$ .

Use the chain rule on $y^2$ with $y$ as the inside function.

The derivative of the inside is $\frac {dy}{dx}$

$6x + 10y\frac {dy}{dx} = 0$

Solve for $\frac {dy}{dx}$ by moving the terms with $\frac {dy}{dx}$ to one side and the rest of the terms to the other side.

If $3x^2 + 5y^2 = 11$ then $\frac {dy}{dx}$ is $\answer {-\frac {3x}{5y}}$

(problem 7b) Use the result of the problem above to find the equation of the tangent line to the ellipse $3x^2 + 5y^2 = 11$ at the point $(\sqrt 2,1).$

$y-y_0 = m(x-x_0)$

$m = \frac {dy}{dx}\bigg |_{(\sqrt 2, 1)}$

The equation of the tangent line is $y = \answer {-\frac {3\sqrt 2}{5} x + \frac {11}{5}}.$

example 8 Find $\frac {dy}{dx}$ if $x^3 + y^3 =6xy$ .
We assume that $y$ is a function of $x$ and we differentiate both sides with respect to $x$ : $\frac {d}{dx}(x^3 + y^3) = \frac {d}{dx} (6xy)$ $\frac {d}{dx}(x^3) + \frac {d}{dx}(y^3) = 6x\frac {dy}{dx} + 6y$ $3x^2 + 3y^2\frac {dy}{dx} = 6x\frac {dy}{dx} + 6y.$ We can now solve for $\displaystyle {\frac {dy}{dx}}$ algebraically: $3y^2\frac {dy}{dx} - 6x\frac {dy}{dx} = 6y-3x^2$ $(3y^2- 6x)\frac {dy}{dx} = 6y-3x^2$ $\frac {dy}{dx} = \frac {6y-3x^2}{3y^2- 6x} = \frac {2y-x^2}{y^2- 2x}.$

Here is a video of Example 3

(problem 8a) Use the result of the example above to find the equation of the tangent line to the curve $x^3 + y^3 = 6xy$ at the point $(3,3).$

$y-y_0 = m(x-x_0)$

$m = \frac {dy}{dx}\bigg |_{(3,3)}$

The equation of the tangent line is $y = \answer {-x + 6}.$

(problem 8b) Find $\frac {dy}{dx}$ if $x^5 + y^5 = 3x^2y$

Consider $y$ to be a function of $x$ .

Differentiate both sides with respect to $x$ .

Use the chain rule on $y^5$ with $y$ as the inside function.

The derivative of the inside is $\frac {dy}{dx}$

Use the product rule on the right hand side.

$5x^4 + 5y^4\frac {dy}{dx} = 3x^2\frac {dy}{dx} + 6xy$

Solve for $\frac {dy}{dx}$ by moving the terms with $\frac {dy}{dx}$ to one side and the rest of the terms to the other side.

If $x^5 + y^5 = 3x^2y$ then $\frac {dy}{dx}$ is $\answer {\frac {6xy - 5x^4}{5y^4 - 3x^2}}$

example 9 Find $\frac {dy}{dx}$ if $x^2(x^2 + y^2) =y^2$ .
We assume that $y$ is a function of $x$ and we differentiate both sides with respect to $x$ : $\frac {d}{dx}[x^2(x^2 + y^2)] = \frac {d}{dx} (y^2 )$ $\frac {d}{dx}(x^4 + x^2y^2) = 2y\frac {dy}{dx}$ $\frac {d}{dx}(x^4) + \frac {d}{dx}(x^2y^2) = 2y\frac {dy}{dx}$ $4x^3 + 2x^2y\frac {dy}{dx} + 2xy^2 = 2y\frac {dy}{dx}.$

We can now solve for $\displaystyle {\frac {dy}{dx}}$ algebraically: $2x^2y\frac {dy}{dx} - 2y\frac {dy}{dx}= -4x^3 - 2xy^2$ $(2x^2y- 2y)\frac {dy}{dx}= -4x^3 - 2xy^2$ $\frac {dy}{dx}= \frac {-4x^3 - 2xy^2}{2x^2y- 2y}$ $\frac {dy}{dx}= -\frac {2x^3 + xy^2}{x^2y- y}= \frac {2x^3 + xy^2}{y - x^2y}.$

Here is a video of Example 4

(problem 9) Find $\frac {dy}{dx}$ if $x\sin (y) - y\cos (x) = xy$

Consider $y$ to be a function of $x$ .

Differentiate both sides with respect to $x$ .

Use the chain rule on $\sin (y)$ with $y$ as the inside function.

The derivative of the inside is $\frac {dy}{dx}$

Use the product rule on both terms on the left hand side and the term on the right hand side.

$x\cos (y)\frac {dy}{dx} + \sin (y) + y\sin (x) - \cos (x)\frac {dy}{dx} = x\frac {dy}{dx} + y$

Solve for $\frac {dy}{dx}$ by moving the terms with $\frac {dy}{dx}$ to one side and the rest of the terms to the other side.

If $x\sin (y) - y\cos (x) = xy$ then $\frac {dy}{dx}$ is $\answer {\frac {y - \sin (y) - y\sin (x)}{x\cos (y) -\cos (x) -x}}$

example 10 Find $\frac {dy}{dx}$ if $(x^2 + y^2)^2 =50xy$ .
We assume that $y$ is a function of $x$ and we differentiate both sides with respect to $x$ : $\frac {d}{dx}[(x^2 + y^2)^2] = \frac {d}{dx} (50xy )$ $\frac {d}{dx}(x^4 + 2x^2y^2 + y^4) = 50x\frac {dy}{dx} + 50y$ $\frac {d}{dx}(x^4) + \frac {d}{dx}(2x^2y^2) + \frac {d}{dx}(y^4) = 50x\frac {dy}{dx} + 50y$ $4x^3 + 4x^2y\frac {dy}{dx} + 4xy^2 + 4y^3 \frac {dy}{dx} = 50x\frac {dy}{dx} + 50y.$

We can now solve for $\displaystyle {\frac {dy}{dx}}$ algebraically: $4x^2y\frac {dy}{dx} + 4y^3 \frac {dy}{dx} - 50x\frac {dy}{dx} = 50y - 4x^3 - 4xy^2$ $(4x^2y + 4y^3 - 50x)\frac {dy}{dx} = 50y - 4x^3 - 4xy^2$ $\frac {dy}{dx} = \frac {50y - 4x^3 - 4xy^2}{4x^2y + 4y^3 - 50x}= \frac {25y - 2x^3 - 2xy^2}{2x^2y + 2y^3 - 25x}.$

Here is a video of Example 5

example 11 Find $\frac {dy}{dx}$ if $\frac {1}{x} + \frac {1}{y} =e^{2y}$ .
We assume that $y$ is a function of $x$ and we differentiate both sides with respect to $x$ : $\frac {d}{dx}\big (\frac {1}{x} + \frac {1}{y}\big ) = \frac {d}{dx} (e^{2y} )$ $\frac {d}{dx}\big (\frac {1}{x}\big ) + \frac {d}{dx}\big (\frac {1}{y}\big ) = \frac {d}{dx} (e^{2y} )$ $-\frac {1}{x^2} - \frac {1}{y^2} \ \frac {dy}{dx} = 2e^{2y} \ \frac {dy}{dx}.$ We can now solve for $\displaystyle {\frac {dy}{dx}}$ algebraically:

$-\frac {1}{x^2} = 2e^{2y} \ \frac {dy}{dx}+ \frac {1}{y^2}\ \frac {dy}{dx}$ $-\frac {1}{x^2} = \Big (2e^{2y}+ \frac {1}{y^2}\Big ) \ \frac {dy}{dx}$ $-\frac {y^2}{x^2} = (2y^2e^{2y}+ 1) \ \frac {dy}{dx}$ $\frac {dy}{dx} = -\frac {y^2}{x^2 (1+ 2y^2e^{2y})}.$

Here is a video of Example 6

(problem 11) Find $\frac {dy}{dx}$ if $e^{xy} + x^2y^2 = 1$

Consider $y$ to be a function of $x$ .

Differentiate both sides with respect to $x$ .

Use the chain rule on $e^{xy}$ with $xy$ as the inside function.

The derivative of the inside is $x\frac {dy}{dx} + y$ by the product rule.

Use the product rule on the $x^2y^2$ term.

$e^{xy} \left ( x\frac {dy}{dx} + y\right ) + 2x^2y \frac {dy}{dx} + 2xy^2 = 0$

Solve for $\frac {dy}{dx}$ by moving the terms with $\frac {dy}{dx}$ to one side and the rest of the terms to the other side.

If $e^{xy} + x^2y^2 = 1$ then $\frac {dy}{dx}$ is $\answer {-\frac {ye^{xy} +2xy^2}{xe^{xy} + 2x^2y}}$

Here are some detailed, lecture style videos on implicit differentiation: