# 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:

$y = \sin x,\;\;\;y = {x^2} + 2x + 5,\;\;\;y = \ln \cos x.$

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

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

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

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

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:

1. 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.
2. 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 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:

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

### 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 $$y' = - {\frac{{{x^3}}}{{{y^3}}}}$$. At the point $$\left( {1,1} \right)$$ we have $$y'\left( 1 \right) = - 1.$$ Hence, the equation of the tangent line is given by

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

### 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.

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

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

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

### Example 9.

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

Solution.

Differentiate both sides term-by-term with respect to $$x:$$

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

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

$6{y^2}y^\prime + y^\prime{x^2} = - \left( {3{x^2} + 2yx} \right),\;\; \Rightarrow y^\prime\left( {{x^2} + 6{y^2}} \right) = - \left( {3{x^2} + 2yx} \right),\; \Rightarrow y^\prime = - \frac{{3{x^2} + 2yx}}{{{x^2} + 6{y^2}}}.$

### Example 10.

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

Solution.

We differentiate this equation with respect to $$x$$ and solve for $$y^\prime:$$

$\left( {{x^5}} \right)^\prime + \left( {{y^5}} \right)^\prime - \left( {2x} \right)^\prime + \left( {2y} \right)^\prime = 0^\prime,\;\; \Rightarrow 5{x^4} + 5{y^4}y^\prime - 2 + 2y^\prime = 0,\;\; \Rightarrow \left( {5{y^4} + 2} \right)y^\prime = 2 - 5{x^4},\;\; \Rightarrow y^\prime = \frac{{2 - 5{x^4}}}{{2 + 5{y^4}}}.$

Substitute the coordinates $$x = 0,$$ $$y = 0:$$

$y^\prime\left( {0,0} \right) = \frac{{2 - 5 \cdot {0^4}}}{{2 + 5 \cdot {0^4}}} = \frac{2}{2} = 1.$

See more problems on Page 2.