2.9 Derivatives of Inverse Trig Functions

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In this section we compute derivatives involving $\tan ^{-1}(x), \sin ^{-1}(x)$ and $\cos ^{-1}(x)$ .

We begin by computing the derivative of the inverse trigonometric function $f(x) = \tan ^{-1}(x)$ . The following Pythagorean trigonometric identity will be needed:

$1 + \tan ^2(\theta ) = \sec ^2(\theta ).$ This identity follows from $\cos ^2(\theta ) + \sin ^2(\theta ) = 1$ by dividing both sides by $\cos ^2(\theta )$ .

We begin the derivation by using the fact that $\tan (x)$ and $\tan ^{-1}(x)$ are inverse functions, so that:

$\tan (\tan ^{-1}(x)) = x.$

We differentiate both sides of this equation with respect to $x$ :

$\dd {x} \tan (\tan ^{-1}(x)) = \ddx x.$

Using the Chain Rule on the left side gives:

$\sec ^2(\tan ^{-1}(x)) \ddx \left [\tan ^{-1}(x)\right ] = 1.$

Now, we can solve this for the derivative of $tan^{-1}(x)$ :

$\ddx \left [\tan ^{-1}(x)\right ] = \frac {1}{\sec ^2\left (\tan ^{-1}(x)\right )}.$

Next, we use the Pythagorean Identity:

$\ddx \left [\tan ^{-1}(x)\right ] = \frac {1}{1 + \tan ^2\left (\tan ^{-1}(x)\right )}.$

Finally, using the property of inverse functions:

$\ddx \left [\tan ^{-1}(x)\right ] = \frac {1}{1 + x^2}.$

Similar arguments can show that:

$\ddx \left [\sin ^{-1}(x)\right ] = \frac {1}{\sqrt {1 - x^2}} \;\;\text { and } \;\;\ddx \left [\cos ^{-1}(x)\right ] = -\frac {1}{\sqrt {1 - x^2}}.$

Below is a graph of $f(x) = \tan ^{-1}(x)$ (in blue) and its derivative, $f'(x) = \tfrac {1}{1+x^2}$ (in purple). Notice that the slope of the tangent line to $f(x)$ (in red) is the height of the corresponding point on $f'(x)$ . Use it to see a graphical representation of the answers to the problem above.

Find the equation of the tangent line to the graph of $f(x) = \tan ^{-1}(x)$ at $x=0.$
The point of tangency is $(0, f(0)) = (0, 0)$ since $\tan ^{-1}(0) = 0$ . The slope of the tangent line is given by $m = f'(0) = \tfrac {1}{1+0^2} = 1$ . To create the equation of the line we use the Point slope form: $y-y_0 = m(x-x_0)$ and we get $y-0 = 1(x-0)$ or $y=x$ .

Find the equation of the tangent line to the graph of $f(x) = \tan ^{-1}(x)$ at $x=1.$

The point of tangency is $(1, f(1))$

Use the derivative to find the slope, $m$

Point slope form: $y-y_0 = m(x-x_0)$

The equation of the tangent line is $y = \answer {x/2 + \pi /4 - 1/2}$

Find the equation of the tangent line to the graph of $f(x) = \sin ^{-1}(x)$ at $x=0.$

The point of tangency is $(0, f(0))$

Use the derivative to find the slope, $m$

Point slope form: $y-y_0 = m(x-x_0)$

The equation of the tangent line is $y = \answer {x}$

Find the equation of the tangent line to the graph of $f(x) = \cos ^{-1}(x)$ at $x=0.$

The point of tangency is $(0, f(0))$

Use the derivative to find the slope, $m$

Point slope form: $y-y_0 = m(x-x_0)$

The equation of the tangent line is $y = \answer {-x + \pi /2}$

Find $h'(x)$ if $h(x) = (1+x^2)\tan ^{-1}(x)$ .
We write $h(x)$ as a product: $h(x) = f(x)\cdot g(x)$ with $f(x) = 1+x^2 \quad \text {and} \quad g(x) = \tan ^{-1}(x).$ To use the product rule we need the derivatives: $f'(x) = 2x \quad \text {and} \quad g'(x) = \frac {1}{1+x^2}.$ We can now write the derivative: $\begin{align*} h'(x) &= f'(x)g(x) + f(x)g'(x) \\ &= 2x\tan^{-1}(x) + (1+x^2)\frac{1}{1+x^2} \\ &= 2x\tan^{-1}(x) + 1. \end{align*}$

Here is a video of the preceding example

Compute $\frac {d}{dx} x^3 \tan ^{-1}(x)$

Use the Product Rule with $x^3$ and $\tan ^{-1}(x)$ .

$(fg)' = f'g+g'f$ .

The derivative of $x^3$ is $3x^2$

The derivative of $\tan ^{-1}(x)$ is $\frac {1}{1+x^2}$

The derivative of $x^3\tan ^{-1}(x)$ with respect to $x$ is $\answer [given]{3x^2 \tan ^{-1}(x) + \frac {x^3}{1+x^2}}$

Compute $\frac {d}{dx} (1-x^2) \sin ^{-1}(x)$

Use the Product Rule with $1-x^2$ and $\sin ^{-1}(x)$ .

$(fg)' = f'g+g'f$ .

The derivative of $1 - x^2$ is $-2x$

The derivative of $\sin ^{-1}(x)$ is $\frac {1}{\sqrt {1-x^2}}$

$(1-x^2) \frac {1}{\sqrt {1-x^2}} = \sqrt {1-x^2}$

The derivative of $(1-x^2)\sin ^{-1}(x)$ with respect to $x$ is $\answer [given]{-2x \sin ^{-1}(x) + \sqrt {1-x^2}}$

The Chain Rule versions of these formulas are:

$\ddx \left [\tan ^{-1}(g(x))\right ] = \frac {g'(x)}{1 + g^2(x)},$ $\ddx \left [\sin ^{-1}(g(x))\right ] = \frac {g'(x)}{\sqrt {1 - g^2(x)}}$

and

$\ddx \left [\cos ^{-1}(g(x))\right ] = -\frac {g'(x)}{\sqrt {1 - g^2(x)}}.$

Find $h'(x)$ if $h(x) = \tan ^{-1}(4x)$ .
We write $h(x)$ as a composition: $h(x)=f(g(x))$ with $g(x) = 4x \quad \text {and} \quad f(x) = \tan ^{-1}(x).$ To find $h'(x)$ we need $f'(g(x))$ and $g'(x)$ . First, $g'(x) = 4; \quad \text {next}$ $f'(x) = \frac {1}{1+x^2} , \quad \text {so}$ $f'(g(x)) = \frac {1}{1+g^2(x)} =\frac {1}{1+(4x)^2}=\frac {1}{1+16x^2}.$ By the chain rule, $h'(x) = f'(g(x))g'(x) = \frac {1}{1+16x^2} \cdot 4 = \frac {4}{1+16x^2}.$

Here is a video of the example

Compute $\frac {d}{dx} \tan ^{-1}(3x)$

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

The outside function is $f(x) = \tan ^{-1}(x)$ and the inside function is $g(x) = 3x$ .

Leave the inside in, $f'(g(x))$

Multiply by the derivative of the inside, $g'(x)$

The derivative of $\tan ^{-1}(3x)$ with respect to $x$ is $\answer [given]{\frac {3}{1+(3x)^2}}$

Find the equation of the tangent line to the graph of $f(x) = \cos (x)$ at $x=\pi /2.$

The point of tangency is $(\pi /2, f(\pi /2))$

Use the derivative to find the slope, $m$

Point slope form: $y-y_0 = m(x-x_0)$

The equation of the tangent line is $y = \answer {-x + \pi /2}$