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Implicit Differentiation


Implicit differentiation makes use of the chain rule to differentiate a function which cannot be explicitly expressed in the form $y=f(x)$. Such a function is known as an implicitly defined function.

By the chain rule, the derivative with respect to $x$ of a function $f \left( y(x) \right)$ is given by:

\[\dfrac{\mathrm{d} }{\mathrm{d} x} \Bigl[ f \left( y(x) \right) \Bigr]=\dfrac{\mathrm{d} f}{\mathrm{d} y} \cdot \dfrac{\mathrm{d} y}{\mathrm{d} x}.\]

Implicit differentiation is useful for, among other things, finding tangent and normal lines to a curve that cannot be expressed in the form $y=f(x)$.

Worked Example

Example 1

Find an expression for $\dfrac{\mathrm{d} y}{\mathrm{d} x}$ given \[\sin{y}=y^2+x\cos{y}+\mathrm{e}^{x}.\]


Differentiate both sides with respect to $x$:

\[\dfrac{\mathrm{d} }{\mathrm{d} x}\Bigl[ \sin{y} \Bigr]=\dfrac{\mathrm{d} }{\mathrm{d} x} \Bigl[ y^2+x\cos{y}+\mathrm{e}^x \Bigr].\]

Recall that, by the chain rule, the derivative of a function $f \left(y(x) \right)$ with respect to $x$ is given by $\dfrac{\mathrm{d} }{\mathrm{d} x} \Bigl[ f \left( y(x) \right) \Bigr] = \dfrac{\mathrm{d} f}{\mathrm{d} y} \cdot \dfrac{\mathrm{d} y}{\mathrm{d} x}$.

Hence, the derivative of $\sin{y}$ with respect to $x$ is:

\[\dfrac{\mathrm{d} }{\mathrm{d} x}\Bigl[ \sin(y) \Bigr] = \frac{\mathrm{d} }{\mathrm{d} y} \Bigl[ \sin{y} \Bigr] \cdot \frac{\mathrm{d} y}{\mathrm{d} x} = \cos{y} \cdot \dfrac{\mathrm{d} y}{\mathrm{d} x},\]

and the derivative of $y^2$ with respect to $x$ is:

\[\dfrac{\mathrm{d} }{\mathrm{d} x} \Bigl[ y^2 \Bigr] = \frac{\mathrm{d} }{\mathrm{d} y} \Bigl[ y^2 \Bigr] \cdot \dfrac{\mathrm{d} y}{\mathrm{d} x}=2y \cdot \dfrac{\mathrm{d} y}{\mathrm{d} x}.\]

Since $x\cos{y}$ is a product of two functions, $x$ and $\cos{y}$, it is necessary to use the product rule to find its derivative.

Recall that the derivative of a product $f(x)g(x)$ with respect to $x$ is $\dfrac{\mathrm{d} }{\mathrm{d} x} \Bigl[ f(x)g(x) \Bigr] =\dfrac{\mathrm{d} f}{\mathrm{d} x}g(x)+f(x)\dfrac{\mathrm{d} g}{\mathrm{d} x}.$

Hence, the derivative of $x\cos{y}$ with respect to $x$ is

\begin{align} \dfrac{\mathrm{d} }{\mathrm{d} x} \Bigl[ x\cos{y} \Bigr] &= \dfrac{\mathrm{d} }{\mathrm{d} x} \Bigl[ x \Bigr] \cdot \cos{y}+x \cdot \dfrac{\mathrm{d} }{\mathrm{d} x} \Bigl[ \cos{y} \Bigr] \\ &= 1 \cdot \cos{y} + x \cdot \dfrac{\mathrm{d} }{\mathrm{d} y} \Bigl[ \cos{y} \Bigr] \cdot \dfrac{\mathrm{d} y}{\mathrm{d} x} \\ &=\cos{y} - x \sin (y) \cdot \dfrac{\mathrm{d} y}{\mathrm{d} x}. \end{align}

Finally, the derivative of $ e^x$ with respect to $x$ is

\[\dfrac{\mathrm{d} }{\mathrm{d} x}\Bigl[ \mathrm{e}^x \Bigr]=\mathrm{e}^x.\]


\[\dfrac{\mathrm{d} }{\mathrm{d} x}\Bigl[y^2+x\cos{y}+\mathrm{e}^x\Bigr]=2y\dfrac{\mathrm{d} y}{\mathrm{d} x}+\cos{y}-x\sin{y}\dfrac{\mathrm{d} y}{\mathrm{d} x}+\mathrm{e}^x.\]

The original equation,

\[\dfrac{\mathrm{d} }{\mathrm{d} x}\Bigl[ \sin{y} \Bigr]=\dfrac{\mathrm{d} }{\mathrm{d} x} \Bigl[ y^2+x\cos{y}+\mathrm{e}^x \Bigr]\]

can now be rewritten by substituting the calculated derivatives:

\[\cos{y}\dfrac{\mathrm{d} y}{\mathrm{d} x}=2y\dfrac{\mathrm{d} y}{\mathrm{d} x}+\cos{y}-x\sin{y}\dfrac{\mathrm{d} y}{\mathrm{d} x}+\mathrm{e}^x.\]

This can be rearranged to find an expression for $\dfrac{\mathrm{d} y}{\mathrm{d} x}$.

Collect terms involving $\dfrac{\mathrm{d} y}{\mathrm{d} x}$ on the left hand side, and other terms on the right:

\[\cos{y}\dfrac{\mathrm{d} y}{\mathrm{d} x}-2y\dfrac{\mathrm{d} y}{\mathrm{d} x}+x\sin{y}\dfrac{\mathrm{d} y}{\mathrm{d} x}=\cos{y}+\mathrm{e}^x.\]

Extract $\dfrac{\mathrm{d} y}{\mathrm{d} x}$ as a factor:

\[\dfrac{\mathrm{d} y}{\mathrm{d} x}\left(\cos{y}-2y+x\sin{y}\right)=\cos{y}+\mathrm{e}^x.\]

Finally, divide both sides by $\cos{y}-2y+x\sin{y}$ to obtain the expression for $\dfrac{\mathrm{d} y}{\mathrm{d} x}$:

\[\dfrac{\mathrm{d} y}{\mathrm{d} x}=\dfrac{\cos{y}+\mathrm{e}^x}{\cos{y}-2y+x\sin{y} }.\]

Video Examples

Example 1

Prof. Robin Johnson finds $\dfrac{\mathrm{d} y}{\mathrm{d} x}$ given an implicitly defined function $y^2=y e^{x}+\cos{y}+x-1$.

Example 2

Prof. Robin Johnson uses implicit differentiation to find $\dfrac{\mathrm{d} }{\mathrm{d} x}\left[\arctan{x}\right]$.

Example 3

Prof. Robin Johnson uses implicit differentiation to find the tangent and normal lines at $(0,0)$ to the curve $3y+2x+x^3=2\sin{y}$.


This workbook produced by HELM is a good revision aid, containing key points for revision and many worked examples.

Test Yourself

Test yourself: Numbas test on implicit differentiation

External Resources

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