# Calculus

## Differentiation of Functions # Implicit Differentiation

If a function is described by the equation $$y = f\left( x \right)$$ where the variable $$y$$ is on the left side, and the right side depends only on the independent variable $$x$$, then the function is said to be given explicitly. For example, the following functions are defined explicitly:

${y = \sin x,\;\;\;}\kern-0.3pt {y = {x^2} + 2x + 5,\;\;\;}\kern-0.0pt {y = \ln \cos x.}$

In many problems, however, the function can be defined in implicit form, that is by the equation

$F\left( {x,y} \right) = 0.$

Of course, any explicit function can be written in an implicit form. So the above functions can be represented as

${y – \sin x = 0,\;\;\;}\kern-0.0pt {y – {x^2} – 2x – 5 = 0,\;\;\;}\kern-0.0pt {y – \ln \cos x = 0.}$

The inverse transformation cannot be always performed. There are often functions defined by an implicit equation that cannot be resolved with respect to the variable $$y.$$ Examples of such implicit functions are

${{x^3} + {y^3} – 3{x^2}{y^5} = 0,\;\;\;}\kern-0.3pt {\frac{{x – y}}{{\sqrt {{x^2} + {y^2}} }} – 4x{y^2} = 0,\;\;\;}\kern-0.3pt {xy – \sin \left( {x + y} \right) = 0.}$

The good news is that we do not need to convert an implicitly defined function into an explicit form to find the derivative $$y’\left( x \right).$$ If $$y$$ is defined implicitly as a function of $$x$$ by an equation $$F\left( {x,y} \right) = 0,$$ we proceed as follows:

• Differentiate both sides of the equation with respect to $$x$$, assuming that $$y$$ is a differentiable function of $$x$$ and using the chain rule. The derivative of zero (in the right side) will also be equal to zero.
Note: If the right side is different from zero, that is the implicit equation has the form
$f\left( {x,y} \right) = g\left( {x,y} \right),$
then we differentiate the left and right side of the equation.
• Solve the resulting equation for the derivative $$y’\left( x \right)$$.

In the examples below find the derivative of the implicit function.

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Find the derivative of the function given by the equation $${y^2} = 2px,$$ where $$p$$ is a parameter.

### Example 2

Differentiate implicitly the function $$y\left( x \right)$$ given by the equation $$y = \cos \left( {x + y} \right).$$

### Example 3

Calculate the derivative at the point $$\left( {0,0} \right)$$ of the function given by the equation $$x = y – 2\sin y.$$

### Example 4

Find the equation of the tangent line to the curve $${x^4} + {y^4} = 2$$ at the point $$\left( {1,1} \right).$$

### Example 5

Calculate the derivative of the function $$y\left( x \right)$$ given by the equation $${x^2} + 2xy + 2{y^2} = 1$$ under condition $$y = 1.$$

### Example 6

Given the equation of a circle $${x^2} + {y^2} = {r^2}$$ of radius $$r$$ centered at the origin. Find the derivative $$y’\left( x \right).$$

### Example 7

${x^2} + {y^2} – 2x – 4y = 4$

### Example 8

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

### Example 9

${x^3} + 2{y^3} + y{x^2} = 3$

### Example 10

Calculate the derivative at the point $$\left( {0,0} \right)$$ of the function given by the equation ${x^5} + {y^5} – 2x + 2y = 0.$

### 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 $${x^{\large\frac{2}{3}\normalsize}} + {y^{\large\frac{2}{3}\normalsize}} = {a^{\large\frac{2}{3}\normalsize}}.$$

### 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

${{2^{\large\frac{x}{y}\normalsize}} = \frac{{{x^2}}}{{{y^2}}}\;\;}\kern-0.3pt{\left( {y \ne 0} \right).}$

### Example 21

${{x^2} + y + \ln \left( {x + y} \right) = 0,\;\;\;}\kern-0.3pt{\left( {x + y \gt 0} \right).}$

### 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

${\frac{y}{x} = \ln \left( {xy} \right)\;\;\;}\kern-0.3pt{\left( {xy \gt 0} \right).}$

### 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 1.

Find the derivative of the function given by the equation $${y^2} = 2px,$$ where $$p$$ is a parameter.

Solution.

This equation is the canonical equation of a parabola. Differentiating the left and right sides with respect to $$x$$, we have:

${{\left( {{y^2}} \right)^\prime } = {\left( {2px} \right)^\prime },\;\;}\Rightarrow {2yy’ = 2p,\;\;}\Rightarrow {y’ = \frac{p}{y},\;\;\;}\kern-0.3pt {\text{where}\;\;y \ne 0.}$

### Example 2.

Differentiate implicitly the function $$y\left( x \right)$$ given by the equation $$y = \cos \left( {x + y} \right).$$

Solution.

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

${\frac{{dy}}{{dx}} = \frac{d}{{dx}}\cos \left( {x + y} \right),\;\;}\Rightarrow {y’ = – \sin \left( {x + y} \right) \cdot \left( {1 + y’} \right),}\Rightarrow {y’ = – \sin \left( {x + y} \right) }-{ y’\sin \left( {x + y} \right),}\Rightarrow {y’\left( {1 + \sin \left( {x + y} \right)} \right) }={ – \sin \left( {x + y} \right),}$

which results in

${y’ }={ – \frac{{\sin \left( {x + y} \right)}}{{1 + \sin \left( {x + y} \right)}}.}$

### Example 3.

Calculate the derivative at the point $$\left( {0,0} \right)$$ of the function given by the equation $$x = y – 2\sin y.$$

Solution.

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

${x^\prime = y^\prime – \left( {2\sin y} \right)^\prime,}\;\; \Rightarrow {1 = y^\prime – 2\cos y \cdot y^\prime,}\;\; \Rightarrow {y^\prime = \frac{1}{{1 – 2\cos y}}.}$

Substitute the coordinates $$\left( {0,0} \right):$$

${y^\prime\left( {0,0} \right) = \frac{1}{{1 – 2\cos 0}} }={ \frac{1}{{1 – 2 \cdot 1}} }={ – 1.}$

### Example 4.

Find the equation of the tangent line to the curve $${x^4} + {y^4} = 2$$ at the point $$\left( {1,1} \right).$$

Solution.

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

${\frac{d}{{dx}}\left( {{x^4} + {y^4}} \right) = \frac{d}{{dx}}\left( 2 \right),\;\;}\Rightarrow {4{x^3} + 4{y^3}y’ = 0,\;\;}\Rightarrow {{x^3} + {y^3}y’ = 0.}$

Then $$y’ = – {\large\frac{{{x^3}}}{{{y^3}}}\normalsize}$$. At the point $$\left( {1,1} \right)$$ we have $$y’\left( 1 \right) = – 1.$$ Hence, the equation of the tangent line is given by

${\frac{{x – 1}}{{y – 1}} = – 1\;\;\;}\kern-0.3pt {\text{or}\;\;x + y = 2.}$

### Example 5.

Calculate the derivative of the function $$y\left( x \right)$$ given by the equation $${x^2} + 2xy + 2{y^2} = 1$$ under condition $$y = 1.$$

Solution.

We differentiate both sides of the equation implicitly with respect to $$x$$ (we consider the left side as a composite function and use the chain rule):

${\frac{d}{{dx}}\left( {{x^2} + 2xy + 2{y^2}} \right) = \frac{d}{{dx}}\left( 1 \right),}\Rightarrow {2x + 2\left( {y + xy’} \right) + 4yy’ = 0,}\Rightarrow {x + y + xy’ + 2yy’ = 0.}$

When $$y = 1,$$ the original equation becomes

${{x^2} + 2x + 2 = 1,\;\;}\Rightarrow {{x^2} + 2x + 1 = 0,\;\;}\Rightarrow {{\left( {x + 1} \right)^2} = 0,\;\;}\Rightarrow {x = – 1.}$

Substituting the values $$x = -1$$ and $$y = 1$$, we obtain:

${ – 1 + 1 – y’ + 2y’ }={ 0.}$

It follows from here that $$y’ = 0$$ at $$y = 1.$$

### Example 6.

Given the equation of a circle $${x^2} + {y^2} = {r^2}$$ of radius $$r$$ centered at the origin. Find the derivative $$y’\left( x \right).$$

Solution.

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

${\frac{d}{{dx}}\left( {{x^2} + {y^2}} \right) = \frac{d}{{dx}}\left( {{r^2}} \right),\;\;}\Rightarrow {2x + 2yy’ = 0,\;\;}\Rightarrow {x + yy’ = 0,\;\;}\Rightarrow {yy’ = – x,\;\;}\Rightarrow {y’ = – \frac{x}{y}.}$

In this case, we can solve for $$y$$ directly from the equation for the upper half-circle: $$y = + \sqrt {{r^2} – {x^2}} .$$ So we get

$y’ = – \frac{x}{{\sqrt {{r^2} – {x^2}} }}.$

### Example 7.

${x^2} + {y^2} – 2x – 4y = 4$

Solution.

We take the derivative of each term treating $$y$$ as a function of $$x.$$

${\left( {{x^2}} \right)^\prime + \left( {{y^2}} \right)^\prime – \left( {2x} \right)^\prime – \left( {4y} \right)^\prime = 4^\prime,}\;\; \Rightarrow {2x + 2yy^\prime – 2 – 4y^\prime = 0.}$

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

${2yy^\prime – 4y^\prime = 2 – 2x,}\;\; \Rightarrow {yy^\prime – 2y^\prime = 1 – x,}\;\; \Rightarrow {y^\prime\left( {y – 2} \right) = 1 – x,}\;\; \Rightarrow {y^\prime = \frac{{1 – x}}{{y – 2}}.}$

### Example 8.

Calculate the derivative of the function given by the equation $${x^3} + {y^3} = 3xy.$$

Solution.

We differentiate the left and right sides of the equation with respect to $$x$$ considering $$y$$ as a composite function of $$x:$$

${{\left( {{x^3} + {y^3}} \right)^\prime } = {\left( {3xy} \right)^\prime },\;\;}\Rightarrow {3{x^2} + 3{y^2}y’ = {\left( {3x} \right)^\prime }y + 3xy’,}\Rightarrow {3{x^2} + 3{y^2}y’ = 3y + 3xy’.}$

From this relation we find $$y’:$$

${{y^2}y’ – xy’ = y – {x^2},\;\;}\Rightarrow {y’\left( {{y^2} – x} \right) = y – {x^2},\;\;}\Rightarrow {y’ = \frac{{y – {x^2}}}{{{y^2} – x}}.}$

This derivative exists provided

${{y^2} – x \ne 0\;\;}\kern0pt{\text{or}\;\;y \ne \pm \sqrt x .}$

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Problems 9-28