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# Calculus

Differentiation of Functions

# Implicit Differentiation

Page 1
Problems 1-4
Page 2
Problems 5-20

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, i.e. 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 $$y.$$ For example, it is impossible to obtain the dependence $$y\left( x \right)$$ in explicit form for the following functions:
${{x^3} + {y^3} – 3{x^2}{y^5} = 0,\;\;\;}\kern-0.3pt {\frac{{x – y}}{{\sqrt {{x^2} + {y^2}} }} – 4x{y^2} = 0,\;\;\;}\kern-0.3pt {xy – \sin \left( {x + y} \right) = 0.}$ The good news is that we do not need to convert an implicitly defined function into an explicit form to find the derivative $$y’\left( x \right).$$ To do this, it is enough to perform the following steps knowing the equation $$F\left( {x,y} \right) = 0:$$

• 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, i.e. 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 $$y’\left( x \right)$$.

The described algorithm for finding the derivative of an implicit function is used in the examples below.

## Solved Problems

Click on problem description to see 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

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

### ✓Example 4

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 5

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 6

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

### ✓Example 7

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

### ✓Example 8

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

### ✓Example 9

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 10

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 11

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

### ✓Example 12

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

### ✓Example 13

$y = \sin \left( {x – y} \right)$

### ✓Example 14

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

### ✓Example 15

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

### ✓Example 16

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

### ✓Example 17

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 18

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

### ✓Example 19

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

### ✓Example 20

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

### 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.

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 4.

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 $$y = 1.$$

Page 1
Problems 1-4
Page 2
Problems 5-20