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