Calculus

Differentiation of Functions

Differentiation Logo

Implicit Differentiation

If a function is described by the equation y = f (x) 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

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.

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