2.13 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

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

Compute $\frac {d}{dx} y^3$

Notice the mismatched letters, $x$ and $y$ , so this is not an ordinary derivative.

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

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

The Chain Rule says $\dfrac {d}{dx} f(g(x)) = f'(g(x)) g'(x).$

The derivative of the inside is written as $\dfrac {dy}{dx}$ or $y'$ .

We have $\frac {d}{dx} y^3 = 3y^2 \dfrac {dy}{dx}.$

The derivative of $y^3$ with respect to $x$ is $\answer [given]{3y^2 \frac {dy}{dx}}$

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

Compute $\frac {d}{dx} \cos (y)$

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

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

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

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

The derivative of $\cos (y)$ with respect to $x$ is $\answer [given]{-\sin (y) \frac {dy}{dx}}$

Compute $\frac {d}{dx} \sec (y)$

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.

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

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

The derivative of $\sec (y)$ with respect to $x$ is $\answer [given]{\sec (y)\tan (y) \frac {dy}{dx}}$

Compute $\frac {d}{dx} e^y$

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

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

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

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

The derivative of $e^y$ with respect to $x$ is $\answer [given]{e^y \frac {dy}{dx}}$

Compute $\frac {d}{dx} 3^y$

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

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

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

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

The derivative of $3^y$ with respect to $x$ is $\answer [given]{3^y \ln (3) \frac {dy}{dx}}$

Compute $\frac {d}{dx} \ln (y)$

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

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

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

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

The derivative of $\ln (y)$ with respect to $x$ is $\answer [given]{\frac {1}{y} \frac {dy}{dx}}$

Compute $\frac {d}{dx} \log (y)$

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

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

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

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

The derivative of $\log (y)$ with respect to $x$ is $\answer [given]{\frac {1}{y \ln (10)} \frac {dy}{dx}}$

Compute $\frac {d}{dx} \tan ^{-1}(y)$

Consider y to be a function of x.

The derivative of $\tan ^{-1}(x)$ with respect to x is $\frac {1}{1+x^2}$ .

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

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

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

The derivative of $\tan ^{-1}(y)$ with respect to $x$ is $\answer [given]{\frac {1}{1+y^2} \frac {dy}{dx}}$

Find $\frac {d}{dx} \big (\frac {1}{y}\big )$ 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

Compute $\frac {d}{dx} \left ( \frac {1}{y^2} \right )$

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

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

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

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

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

The derivative of $\frac {1}{y^2}$ with respect to $x$ is $\answer [given]{-\frac {2}{y^3} \frac {dy}{dx}}$

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

Compute $\frac {d}{dx} 3x^2y$

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

Use the product rule with $3x^2$ and $y$ .

The derivative of $3x^2y$ with respect to $x$ is $\answer [given]{3x^2 \frac {dy}{dx} + 6xy}$

Find $\frac {d}{dx} (x^2y^3)$ 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 $f(x) = x^2$ 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

Compute $\frac {d}{dx} x^3y^4$

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

Use the product rule with $x^3$ and $y^4$ .

Use the chain rule on $y^4$ with $y$ as the inside function.

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

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

The derivative of $x^3 y^4$ with respect to $x$ is $\answer [given]{4x^3 y^3 \frac {dy}{dx} + 3x^2y^4}$

Compute $\frac {d}{dx} x^2e^y$

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

Use the product rule with $x^2$ and $e^y$ .

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

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

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

The derivative of $x^2 e^y$ with respect to $x$ is $\answer [given]{x^2 e^y \frac {dy}{dx} + 2xe^y}$

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

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

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 $y = \answer {\frac 34 x+\frac 54}.$

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 $\answer [given]{-\frac {x}{y}}$

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

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 $\answer [given]{-\frac {3x}{5y}}$

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 $y = \answer {-\frac {3\sqrt 2}{5} x + \frac {11}{5}}.$

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

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 $y = \answer {-x + 6}.$

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 $\answer [given]{\frac {6xy - 5x^4}{5y^4 - 3x^2}}$

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

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 $\answer [given]{\frac {y - \sin (y) - y\sin (x)}{x\cos (y) -\cos (x) -x}}$

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

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

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 $\answer [given]{-\frac {ye^{xy} +2xy^2}{xe^{xy} + 2x^2y}}$

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