### Differentiation using Inverse Functions

#### Definition

Differentiation using inverse functions is a technique for finding derivatives. It involves finding the inverse of the function to be differentiated and then applying implicit differentiation. This technique is particularly useful for finding the derivatives of the inverse trigonometric functions $\arccos{x}$, $\arcsin{x}$ and $\arctan{x}$.

#### Worked Examples

###### Example 1 - The Derivative of arctan

Find $\dfrac{\mathrm{d {\mathrm{d} x}\Bigl[\arcsin{x}\Bigl]$.

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###### Solution

Let $y=\arcsin{x}$. We require the inverse of this function to proceed with the technique.

By definition, $\arcsin{x}$ is the inverse of $\sin{x}$. Therefore the inverse function of $y=\arcsin{x}$ is:

$\sin{y}=x.$

Differentiating both sides of this equation gives:

$\dfrac{\mathrm{d </div>{\mathrm{d} x}\Bigl[\sin{y}\Bigl]=\dfrac{\mathrm{d}}{\mathrm{d} x}\Bigl[x\Bigl].$

Since the right-hand side of this equation is purely a function of $x$, it can be differentiated immediately. However, it is necessary to apply the chain rule to the left-hand side, since the term to be differentiated is a function of $y(x)$. Applying the chain rule gives:

\begin{align} \dfrac{\mathrm{d}}{\mathrm{d} x}\Bigl[\sin{y}\Bigl] &= \dfrac{\mathrm{d}}{\mathrm{d} x}\Bigl[x\Bigl], \\ \dfrac{\mathrm{d}}{\mathrm{d} y}\Bigl[\sin{y}\Bigl]\dfrac{\mathrm{d} y}{\mathrm{d} x} &= 1, \\ \cos{y}\dfrac{\mathrm{d} y}{\mathrm{d} x} &= 1. \end{align}

Dividing both sides by $\cos{y}$ gives an expression for $\dfrac{\mathrm{d} y}{\mathrm{d} x}$:

$\dfrac{\mathrm{d} y}{\mathrm{d} x}=\dfrac{1}{\cos{y} }.$

This must now be transformed back into an expression in terms of $x$.

Recall the trigonometric identity $\cos^2{y}+\sin^2{y}=1$. Rearranging this gives $\cos{y}=\sqrt{1-\sin^2{y} }$.

Here we have $\sin{y}=x$, so $\cos{y}$ can be expressed in the following form:

$\cos{y}=\sqrt{1-\sin^2{y} }=\sqrt{1-x^2}.$

Substituting this into the expression for $\dfrac{\mathrm{d} y}{\mathrm{d} x}$ gives the derivative of $\arcsin{x}$:

$\dfrac{\mathrm{d} y}{\mathrm{d} x}=\dfrac{\mathrm{d}}{\mathrm{d} x}\Bigl[\arcsin{x}\Bigl]=\dfrac{1}{\sqrt{1-x^2} }.$

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###### Example 2 - The derivative of arctan(x)

Find $\dfrac{\mathrm{d {\mathrm{d} x}\Bigl[\arctan{x}\Bigl]$.

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###### Solution

Let $y=\arctan{x}$. We require the inverse of this function to proceed with the technique.

By definition, $\arctan{x}$ is the inverse of $\tan{x}$. Therefore the inverse function of $y=\arctan{x}$ is:

$\tan{y}=x.$

Differentiating both sides of this equation gives:

$\dfrac{\mathrm{d </div>{\mathrm{d} x}\Bigl[\tan{y}\Bigl]=\dfrac{\mathrm{d}}{\mathrm{d} x}\Bigl[x\Bigl].$

Since the right-hand side of this equation is purely a function of $x$ , it can be differentiated immediately. However, it is necessary to apply the chain rule to the left-hand side, since the term to be differentiated is a function of $y(x)$ . Applying the chain rule gives:

\begin{align} \dfrac{\mathrm{d}}{\mathrm{d} x}\Bigl[\tan{y}\Bigl] &= \dfrac{\mathrm{d}}{\mathrm{d} x}\Bigl[x\Bigl], \\ \dfrac{\mathrm{d}}{\mathrm{d} y}\Bigl[\tan{y}\Bigl]\dfrac{\mathrm{d} y}{\mathrm{d} x} &= 1, \\ \sec^2{y}\dfrac{\mathrm{d} y}{\mathrm{d} x} &= 1. \end{align}

Dividing both sides by $\sec^2{y}$ gives an expression for $\dfrac{\mathrm{d} y}{\mathrm{d} x}$:

$\dfrac{\mathrm{d} y}{\mathrm{d} x}=\dfrac{1}{\sec^2{y} }.$

This must now be transformed back into an expression in terms of $x$.

Recall the trigonometric identity $\sec^2{y}=1+\tan^2{y}$. Using this identity and recalling that $\tan{y}=x$ allows us to express $\dfrac{\mathrm{d} y}{\mathrm{d} x}$ in terms of $x$:

\begin{align} \dfrac{\mathrm{d} y}{\mathrm{d} x} &=\dfrac{1}{\sec^2{y} } \\ &=\dfrac{1}{1+\tan^2{y} } \\ &=\dfrac{1}{1+x^2} \end{align}

Hence the derivative of $\arctan{x}$ is:

$\dfrac{\mathrm{d}}{\mathrm{d} x}\Bigl[\arctan{x}\Bigl]=\dfrac{1}{1+x^2}.$

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#### Video Examples

###### Example 1

Prof. Robin Johnson uses inverse functions to find $\dfrac{\mathrm{d} }{\mathrm{d}x}\bigl[\ln{(x^2-x+1)}\bigl]$.

###### Example 2

Prof. Robin Johnson uses inverse functions to find $\dfrac{\mathrm{d} }{\mathrm{d}x}\Bigl[\ln{\bigl(\sin{(x+x^2)}\bigl)}\Bigl]$.

###### Example 3

Prof. Robin Johnson uses inverse functions to find $\dfrac{\mathrm{d} }{\mathrm{d}x}\Bigl[\arctan{\bigl(\sqrt{1+x}\bigl)}\Bigl]$. #### More Support

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