# Implicit Differentiation

## Solved Problems

Click or tap a problem to see the solution.

### Example 11

$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$

### Example 12

${3^x} + {3^y} = {3^{x + y}}$

### Example 13

Find the derivative of the astroid ${x^{\frac{2}{3}}} + {y^{\frac{2}{3}}} = {a^{\frac{2}{3}}}.$

### Example 14

Find the derivative $$y'\left( x \right)$$ of the function that describes the general equation of a second order curve:

$$A{x^2} + 2Bxy + C{y^2} + 2Dx +$$ $$2Ey + F = 0.$$

### Example 15

Find the value of the derivative $$y'\left( x \right)$$ at $$x = 0$$ for the function given by $${e^y} + xy = e.$$

### Example 16

$x\sin y + y\cos x = 0$

### Example 17

$y = \sin \left( {x - y} \right)$

### Example 18

$y = \sin \left( {x + y} \right)$

### Example 19

${x^2} - \sin \left( {xy} \right) = 0$

### Example 20

${2^{\frac{x}{y}}} = \frac{{{x^2}}}{{{y^2}}}\;\left( {y \ne 0} \right).$

### Example 21

${x^2} + y + \ln \left( {x + y} \right) = 0,$ where $$x + y \gt 0.$$

### Example 22

Find the derivative at the point $$x = 2$$ for the function given by the equation

$${x^2} + {y^2} + 2x - xy +$$ $${ 5y - 2 = 0.}$$

### Example 23

$\frac{y}{x} = \ln \left( {xy} \right),$ where $$xy \gt 0.$$

### Example 24

Calculate the derivative at the point $$\left( {1,1} \right)$$ of the function given by the equation $x - y = \ln \left( {xy} \right),$ where $$xy \gt 0.$$

### Example 25

Find the derivative of the function defined by the equation $x + y = \arctan \left( {xy} \right).$

### Example 26

$x - y + \arctan y = 0$

### Example 27

$x + y = {e^{x - y}}$

### Example 28

${x^y} = {y^x}$

### Example 11.

$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$

Solution.

The written equation is the equation of an ellipse. Differentiating both sides and taking into account that $$y$$ is a function of $$x$$, we have:

${\left( {\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}}} \right)^\prime } = {\left( 1 \right)^\prime },\;\; \Rightarrow \frac{{2x}}{{{a^2}}} + \frac{{2y}}{{{b^2}}}y' = 0,\;\; \Rightarrow y'\frac{y}{{{b^2}}} = - \frac{x}{{{a^2}}},\;\; \Rightarrow y' = - \frac{{{b^2}x}}{{{a^2}y}},\;\;\text{where}\;\;y \ne 0.$

### Example 12.

${3^x} + {3^y} = {3^{x + y}}$

Solution.

Differentiate both sides and solve the resulting equation for $$y':$$

$\left( {{3^x} + {3^y}} \right)^\prime = \left( {{3^{x + y}}} \right)^\prime,\;\; \Rightarrow {3^x}\ln 3 + {3^y}\ln 3 \cdot y' = {3^{x + y}}\ln 3 \cdot {\left( {x + y} \right)^\prime },\;\; \Rightarrow {3^x} + {3^y}y' = {3^{x + y}}\left( {1 + y'} \right),\;\; \Rightarrow {3^x} + {3^y}y' = {3^{x + y}} + {3^{x + y}}y',\;\; \Rightarrow {3^y}y' - {3^{x + y}}y' = {3^{x + y}} - {3^x},\;\; \Rightarrow {3^y}y'\left( {1 - {3^x}} \right) = {3^x}\left( {{3^y} - 1} \right), \;\Rightarrow y' = - \frac{{{3^x}\left( {{3^y} - 1} \right)}}{{{3^y}\left( {{3^x} - 1} \right)}}.$

### Example 13.

Find the derivative of the astroid ${x^{\frac{2}{3}}} + {y^{\frac{2}{3}}} = {a^{\frac{2}{3}}}.$

Solution.

Differentiate the given equation with respect to $$x:$$

$\left( {{x^{\frac{2}{3}}} + {y^{\frac{2}{3}}}} \right)^\prime = \left( {{a^{\frac{2}{3}}}} \right)^\prime,\;\; \Rightarrow \frac{2}{3}{x^{ - \frac{1}{3}}} + \frac{2}{3}{y^{ - \frac{1}{3}}}y' = 0,\;\; \Rightarrow \frac{1}{{\sqrt[3]{x}}} + \frac{{y'}}{{\sqrt[3]{y}}} = 0,\;\; \Rightarrow \frac{{y'}}{{\sqrt[3]{y}}} = - \frac{1}{{\sqrt[3]{x}}},\;\; \Rightarrow y' = - \sqrt[3]{{\frac{y}{x}}},\;\;\text{where}\;\;x \ne 0,\;y \ne 0.$

### Example 14.

Find the derivative $$y'\left( x \right)$$ of the function that describes the general equation of a second order curve:

$$A{x^2} + 2Bxy + C{y^2} + 2Dx +$$ $$2Ey + F = 0.$$

Solution.

We differentiate this equation with respect to $$x$$, taking into account that $$y$$ is a function of $$x:$$

$\left( {A{x^2} + 2Bxy + C{y^2} + 2Dx + 2Ey + F} \right)^\prime = 0,\;\; \Rightarrow 2Ax + 2B\left( {x'y + xy'} \right) + 2Cyy' + 2D + 2Ey' = 0,\;\; \Rightarrow 2Ax + 2By + 2Bxy' + 2Cyy' + 2D + 2Ey' = 0,\;\; \Rightarrow Bxy' + Cyy' + Ey' = - \left( {Ax + By + D} \right),\;\; \Rightarrow y' = - \frac{{Ax + By + D}}{{Bx + Cy + E}}.$

### Example 15.

Find the value of the derivative $$y'\left( x \right)$$ at $$x = 0$$ for the function given by $${e^y} + xy = e.$$

Solution.

Differentiating both sides, we obtain the following relationship:

$\left( {{e^y} + xy} \right)^\prime = \left( e \right)^\prime,\;\; \Rightarrow {e^y}y' + x'y + xy' = 0,\;\; \Rightarrow {e^y}y' + y + xy' = 0.$

From here we find the derivative:

$y'\left( {{e^y} + x} \right) = - y,\;\; \Rightarrow y' = - \frac{y}{{{e^y} + x}}.$

Calculate the value of $$y$$ at $$x = 0:$$

${e^0} + 0 \cdot y = e,\;\; \Rightarrow {e^y} = e,\;\; \Rightarrow y = 1.$

Then the derivative at the point $$\left( {0,1} \right)$$ is equal to

$y'\left( {x = 0,y = 1} \right) = - \frac{1}{{{e^1} + 0}} = - \frac{1}{e}.$

### Example 16.

$x\sin y + y\cos x = 0$

Solution.

As usual, we differentiate both sides with respect to $$x:$$

$\left( {x\sin y + y\cos x} \right)^\prime = 0,\;\; \Rightarrow \left( {x\sin y} \right)^\prime + \left( {y\cos x} \right)^\prime = 0.$

Using the product rule for the derivative, we have:

$x'\sin y + x{\left( {\sin y} \right)^\prime } + y'\cos x + y{\left( {\cos x} \right)^\prime } = 0,\;\; \Rightarrow 1 \cdot \sin y + x \cdot \cos y \cdot y' + y' \cdot \cos x + y \cdot \left( { - \sin x} \right) = 0,\; \Rightarrow \sin y + x\cos y \cdot y' + \cos x \cdot y' - y\sin x = 0,\;\; \Rightarrow y'\left( {x\cos y + \cos x} \right) = y\sin x - \sin y,\;\; \Rightarrow y' = \frac{{y\sin x - \sin y}}{{x\cos y + \cos x}}.$

### Example 17.

$y = \sin \left( {x - y} \right)$

Solution.

Differentiate both sides of the equation with respect to $$x:$$

$y' = {\left[ {\sin \left( {x - y} \right)} \right]^\prime },\;\; \Rightarrow y' = \cos \left( {x - y} \right) \cdot {\left( {x - y} \right)^\prime },\;\; \Rightarrow y' = \cos \left( {x - y} \right) \cdot \left( {1 - y'} \right),\;\; \Rightarrow y' = \cos \left( {x - y} \right) - \cos \left( {x - y} \right) \cdot y',\;\; \Rightarrow y'\left[ {1 + \cos \left( {x - y} \right)} \right] = \cos \left( {x - y} \right)\;\; \Rightarrow y' = \frac{{\cos \left( {x - y} \right)}}{{1 + \cos \left( {x - y} \right)}},$

where the derivative $$y'$$ is defined provided that

$1 + \cos \left( {x - y} \right) \ne 0,\;\; \Rightarrow \cos \left( {x - y} \right) \ne - 1,\;\; \Rightarrow x - y \ne \pi + 2\pi n,\;\;n \in \mathbb{Z}.$

### Example 18.

$y = \sin \left( {x + y} \right)$

Solution.

$y' = {\left[ {\sin \left( {x + y} \right)} \right]^\prime },\;\; \Rightarrow y' = \cos \left( {x + y} \right) \cdot {\left( {x + y} \right)^\prime },\;\; \Rightarrow y' = \cos \left( {x + y} \right) \cdot \left( {1 + y'} \right),\;\; \Rightarrow y' = \cos \left( {x + y} \right) + \cos \left( {x + y} \right) \cdot y',\;\; \Rightarrow y'\left[ {1 - \cos \left( {x + y} \right)} \right] = \cos \left( {x + y} \right)\;\; \Rightarrow y' = \frac{{\cos \left( {x + y} \right)}}{{1 - \cos \left( {x + y} \right)}}.$

It is seen that the derivative is defined under condition

$1 - \cos \left( {x + y} \right) \ne 0,\;\; \Rightarrow \cos \left( {x + y} \right) \ne 1,\;\; \Rightarrow x + y \ne 2\pi n,\;\;n \in \mathbb{Z}.$

### Example 19.

${x^2} - \sin \left( {xy} \right) = 0$

We differentiate each term with respect to $$x:$$

$\left( {{x^2}} \right)^\prime - \left( {\sin \left( {xy} \right)} \right)^\prime = 0^\prime,\;\; \Rightarrow 2x - \cos \left( {xy} \right) \cdot \left( {xy} \right)^\prime = 0,\;\; \Rightarrow 2x - \cos \left( {xy} \right) \cdot \left( {y + xy^\prime} \right) = 0,\;\; \Rightarrow 2x - y\cos \left( {xy} \right) - x\cos \left( {xy} \right)y^\prime = 0.$

Solve for $$y^\prime:$$

$x\cos \left( {xy} \right)y^\prime = 2x - y\cos \left( {xy} \right),\;\; \Rightarrow y^\prime = \frac{{2x - y\cos \left( {xy} \right)}}{{x\cos \left( {xy} \right)}},\;\; \Rightarrow y^\prime = \frac{2}{{\cos \left( {xy} \right)}} - \frac{y}{x}.$

### Example 20.

${2^{\frac{x}{y}}} = \frac{{{x^2}}}{{{y^2}}}\;\left( {y \ne 0} \right).$

Solution.

Following the pattern described above, we have:

$\left( {{2^{\frac{x}{y}}}} \right)^\prime = \left( {\frac{{{x^2}}}{{{y^2}}}} \right)^\prime,\;\; \Rightarrow {2^{\frac{x}{y}}}\ln 2 \cdot {\left( {\frac{x}{y}} \right)^\prime } = \frac{{{{\left( {{x^2}} \right)}^\prime }{y^2} - {x^2}{{\left( {{y^2}} \right)}^\prime }}}{{{y^4}}},\;\; \Rightarrow {2^{\frac{x}{y}}}\ln 2 \cdot \frac{{x'y - xy'}}{{{y^2}}} = \frac{{2x \cdot {y^2} - {x^2} \cdot 2y \cdot y'}}{{{y^4}}},\;\; \Rightarrow {2^{\frac{x}{y}}}\ln 2 \cdot \frac{{y - xy'}}{{{y^2}}} = \frac{{2xy\left( {y - xy'} \right)}}{{{y^4}}},\;\; \Rightarrow {2^{\frac{x}{y}}}\ln 2\left( {y - xy'} \right) = \frac{{2x}}{y}\left( {y - xy'} \right),\;\; \Rightarrow \left( {{2^{\frac{x}{y}}}\ln 2 - \frac{{2x}}{y}} \right) \cdot \left( {y - xy'} \right) = 0.$

From the last equation we find the derivative:

$y - xy' = 0,\;\; \Rightarrow y' = \frac{y}{x},\;\;\text{where}\;\;\;x \ne 0.$

Note that this function does not exist at $$y = 0.$$ At the same time, the derivative exists neither at $$y = 0$$ nor at $$x = 0.$$

### Example 21.

${x^2} + y + \ln \left( {x + y} \right) = 0,$ where $$x + y \gt 0.$$

Solution.

Compute the derivative $$y'$$ following the common pattern described above. As a result, we have:

$\left[ {{x^2} + y + \ln \left( {x + y} \right)} \right]^\prime = 0,\;\; \Rightarrow {2x + y' + \frac{1}{{x + y}} \cdot {\left( {x + y} \right)^\prime } = 0,\;\;} \Rightarrow {2x + y' + \frac{1}{{x + y}} \cdot \left( {1 + y'} \right) = 0,\;\;} \Rightarrow {2x + y' + \frac{1}{{x + y}} + \frac{{y'}}{{x + y}} = 0,\;\;} \Rightarrow {y'\left( {1 + \frac{1}{{x + y}}} \right) = - \left( {2x + \frac{1}{{x + y}}} \right),\;\;} \Rightarrow {y' \cdot \frac{{x + y + 1}}{{x + y}} = - \frac{{2{x^2} + 2xy + 1}}{{x + y}},\;\;} \Rightarrow {y' = - \frac{{2{x^2} + 2xy + 1}}{{x + y + 1}}.}$

### Example 22.

Find the derivative at the point $$x = 2$$ for the function given by the equation

$${x^2} + {y^2} + 2x - xy +$$ $${ 5y - 2 = 0.}$$

Solution.

Pre-calculate the value of $$y$$ at $$x = 2:$$

${x^2} + {y^2} + 2x - xy + 5y - 2 = 0,\;\; \Rightarrow {y^2} + 3y + 6 = 0,\;\; \Rightarrow {\left( {y + 3} \right)^2} = 0,\;\; \Rightarrow y = - 3.$

Now we differentiate the given equation with respect to $$x:$$

$\left( {{x^2} + {y^2} + 2x - xy + 5y - 2} \right)^\prime = 0,\;\; \Rightarrow 2x + 2yy' + 2 - \left( {x'y + xy'} \right) + 5y' = 0,\;\; \Rightarrow 2x + 2yy' + 2 - y - xy' + 5y' = 0,\;\; \Rightarrow 2yy' - xy' + 5y' = y - 2x - 2,\;\; \Rightarrow y'\left( {2y - x + 5} \right) = y - 2x - 2,\;\; \Rightarrow y' = \frac{{y - 2x - 2}}{{2y - x + 5}}.$

Substituting the coordinates of the point $$\left( {x = 2,y = - 3} \right),$$ we find the value of the derivative:

$y'\left( {2, - 3} \right) = \frac{{ - 3 - 2 \cdot 2 - 2}}{{2 \cdot \left( { - 3} \right) - 2 + 5}} = \frac{{ - 9}}{{ - 3}} = 3.$

### Example 23.

$\frac{y}{x} = \ln \left( {xy} \right),$ where $$xy \gt 0.$$

Solution.

Using the product, quotient and chain rules, we obtain:

$\left( {\frac{y}{x}} \right)^\prime = \left[ {\ln \left( {xy} \right)} \right]^\prime,\;\; \Rightarrow \frac{{y'x - yx'}}{{{x^2}}} = \frac{1}{{xy}} \cdot {\left( {xy} \right)^\prime },\;\; \Rightarrow \frac{{y'x - y}}{{{x^2}}} = \frac{{x'y + xy'}}{{xy}},\;\; \Rightarrow \frac{{y'x - y}}{x} = \frac{{y + xy'}}{y}.$

Solve this equation for $$y':$$

$y' - \frac{y}{x} = 1 + \frac{x}{y}y',\;\; \Rightarrow y' - \frac{x}{y}y' = 1 + \frac{y}{x},\;\; \Rightarrow y'\left( {1 - \frac{x}{y}} \right) = 1 + \frac{y}{x},\;\; \Rightarrow y' \cdot \frac{{y - x}}{y} = \frac{{y + x}}{x},\;\; \Rightarrow y' = \frac{{y + x}}{x} \cdot \frac{y}{{y - x}},\;\; \Rightarrow y' = \frac{{y\left( {y + x} \right)}}{{x\left( {y - x} \right)}}.$

assuming that $$y \ne x.$$

### Example 24.

Calculate the derivative at the point $$\left( {1,1} \right)$$ of the function given by the equation $x - y = \ln \left( {xy} \right),$ where $$xy \gt 0.$$

Solution.

Differentiate both sides of the equation with respect to $$x$$ and solve for $$y^\prime:$$

$\left( {x - y} \right)^\prime = \left( {\ln \left( {xy} \right)} \right)^\prime,\;\; \Rightarrow 1 - y^\prime = \frac{1}{{xy}} \cdot \left( {y + xy^\prime} \right),\;\; \Rightarrow xy - xyy^\prime = y + xy^\prime,\;\; \Rightarrow xy - y = y^\prime\left( {xy + x} \right),\;\; \Rightarrow y^\prime = \frac{{xy - y}}{{xy + x}}.$

Substitute the coordinates $$\left( {1,1} \right):$$

$y^\prime\left( {1,1} \right) = \frac{{1 \cdot 1 - 1}}{{1 \cdot 1 + 1}} = 0.$

### Example 25.

Find the derivative of the function defined by the equation $x + y = \arctan \left( {xy} \right).$

Solution.

Differentiate both sides of the equation with respect to $$x:$$

${\left( {x + y} \right)^\prime } = {\left[ {\arctan \left( {xy} \right)} \right]^\prime },\;\; \Rightarrow {1 + y' = \frac{1}{{1 + {{\left( {xy} \right)}^2}}} \cdot {\left( {xy} \right)^\prime }.}$

Using the product rule for the derivative, we can write:

$1 + y' = \frac{1}{{1 + {{\left( {xy} \right)}^2}}} \cdot \left( {x'y + xy'} \right),\;\; \Rightarrow 1 + y' = \frac{y}{{1 + {x^2}{y^2}}} + \frac{{xy'}}{{1 + {x^2}{y^2}}}.$

Consequently,

${y' - \frac{{xy'}}{{1 + {x^2}{y^2}}} = \frac{y}{{1 + {x^2}{y^2}}} - 1,\;\;} \Rightarrow {y'\left( {1 - \frac{x}{{1 + {x^2}{y^2}}}} \right) = \frac{{y - 1 - {x^2}{y^2}}}{{1 + {x^2}{y^2}}},\;\;} \Rightarrow {y' \cdot \frac{{{x^2}{y^2} - x + 1}}{{1 + {x^2}{y^2}}} = - \frac{{{x^2}{y^2} - y + 1}}{{1 + {x^2}{y^2}}},\;\;} \Rightarrow {y' = - \frac{{{x^2}{y^2} - y + 1}}{{{x^2}{y^2} - x + 1}}.}$

Note that the derivative $$y'$$ in this example exists provided that

${x^2}{y^2} - x + 1 \ne 0,\;\; \Rightarrow {x^2}{y^2} \ne x - 1,\;\; \Rightarrow {y^2} \ne \frac{{x - 1}}{{{x^2}}},\;\; \Rightarrow y \ne \pm \sqrt {\frac{{x - 1}}{{{x^2}}}},\;\; \Rightarrow y \ne \pm \frac{{\sqrt {x - 1} }}{{\left| x \right|}},\;\; \Rightarrow y \ne \pm \frac{{\sqrt {x - 1} }}{x}\;\;\text{and}\;\;x \gt 1.$

### Example 26.

$x - y + \arctan y = 0$

Solution.

Differentiate both sides with respect to $$x:$$

$x^\prime - y^\prime + \left( {\arctan y} \right)^\prime = 0^\prime,\;\; \Rightarrow 1 - y^\prime + \frac{{y'}}{{1 + {y^2}}} = 0.$

Solve this equation for $$y^\prime:$$

$y^\prime\left( {1 - \frac{1}{{1 + {y^2}}}} \right) = 1,\;\; \Rightarrow y^\prime \cdot \frac{{1 + {y^2} - 1}}{{1 + {y^2}}} = 1,\;\; \Rightarrow \frac{{y^\prime{y^2}}}{{1 + {y^2}}} = 1,\;\; \Rightarrow y^\prime = \frac{{1 + {y^2}}}{{{y^2}}},\;\; \Rightarrow y^\prime = \frac{1}{{{y^2}}} + 1.$

### Example 27.

$x + y = {e^{x - y}}$

Solution.

First we take logarithms of both sides:

$\ln \left( {x + y} \right) = \ln {e^{x - y}},\;\; \Rightarrow \ln \left( {x + y} \right) = x - y.$

Differentiate this equation with respect to $$x:$$

$\left( {\ln \left( {x + y} \right)} \right)^\prime = \left( {x - y} \right)^\prime,\;\; \Rightarrow \frac{1}{{x + y}} \cdot \left( {1 + y^\prime} \right) = 1 - y^\prime,\;\; \Rightarrow 1 + y^\prime = \left( {1 - y^\prime} \right)\left( {x + y} \right),\;\; \Rightarrow 1 + y^\prime = x - xy^\prime + y - yy^\prime,\;\; \Rightarrow y^\prime + xy^\prime + yy^\prime = x + y - 1,\;\; \Rightarrow y^\prime\left( {1 + x + y} \right) = x + y - 1,\;\; \Rightarrow y^\prime = \frac{{x + y - 1}}{{x + y + 1}}.$

### Example 28.

${x^y} = {y^x}$

Solution.

First, we take logarithms of both sides of the equation:

$\ln \left( {{x^y}} \right) = \ln \left( {{y^x}} \right),\;\; \Rightarrow y\ln x = x\ln y.$

Here it is assumed that $$x \gt 0$$ and $$y \gt 0.$$ Moreover, $$x \ne 1$$ and $$y \ne 1$$ as bases of the exponential functions.

Differentiating both sides of the equality with respect to $$x$$ and taking into account that $$y$$ is a function of $$x,$$ we get:

$\left( {y\ln x} \right)^\prime = \left( {x\ln y} \right)^\prime,\;\; \Rightarrow y' \cdot \ln x + y \cdot {\left( {\ln x} \right)^\prime } = x' \cdot \ln y + x \cdot {\left( {\ln y} \right)^\prime},\;\; \Rightarrow y' \cdot \ln x + y \cdot \frac{1}{x} = 1 \cdot \ln y + x \cdot \frac{1}{y} \cdot y',\;\; \Rightarrow y'\left( {\ln x - \frac{x}{y}} \right) = \ln y - \frac{y}{x},\;\; \Rightarrow y' = \frac{{\ln y - \frac{y}{x}}}{{\ln x - \frac{x}{y}}},\;\; \Rightarrow y' = \frac{{y\left( {x\ln y - y} \right)}}{{x\left( {y\ln x - x} \right)}}.$

Note that in addition to the restrictions on the allowed values of $$x$$ indicated above, the derivative has a discontinuity under condition

$y\ln x - x = 0\;\;\text{or}\;\;y = \frac{x}{{\ln x}}.$