Derivatives of Inverse Functions: Fresh Take

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Derivatives of Inverse Functions: Fresh Take

Find the derivative of an inverse function
Identify the derivatives for inverse trig functions like arcsine, arccosine, and arctangent

Derivatives of Various Inverse Functions

The Main Idea

Inverse Function Theorem:
- For invertible and differentiable function f(x): $\dfrac{1}{f^{\prime}(f^{-1}(x))}$
Graphical Interpretation:
- Tangent lines of $f(x)$ and $f^(-1)(x)$ have reciprocal slopes
- Symmetric about $y = x$ line
Extending the Power Rule:
- For positive integer $n$ : $\frac{d}{dx}(x^{1/n}) = \frac{1}{n}x^{(1/n)-1}$
- For positive integer $n$ and any integer m: $\frac{d}{dx}(x^{m/n}) = \frac{m}{n}x^{(m/n)-1}$

Use the inverse function theorem to find the derivative of $g(x)=\dfrac{1}{x+2}.$ Compare the result obtained by differentiating $g(x)$ directly.

g^{'} (x) = - \frac{1}{(x + 2)^{2}}

$g^{\prime}(x)=-\frac{1}{(x+2)^2}$

Use the inverse function theorem to find the derivative of $g(x)=\sqrt[3]{x}$ .

The function $g(x)=\sqrt[3]{x}$ is the inverse of the function $f(x)=x^3$ . Since $g^{\prime}(x)=\frac{1}{f^{\prime}(g(x))}$ , begin by finding $f^{\prime}(x)$ . Thus,

f^{'} (x) = 3 x^{2}

$f^{\prime}(x)=3x^2$ and

f^{'} (g (x)) = 3 (\sqrt[3]{x})^{2} = 3 x^{2 / 3}

$f^{\prime}(g(x))=3(\sqrt[3]{x})^2=3x^{2/3}$

Finally,

g^{'} (x) = \frac{1}{3 x^{2 / 3}} = \frac{1}{3} x^{- 2 / 3}

$g^{\prime}(x)=\frac{1}{3x^{2/3}}=\frac{1}{3}x^{-2/3}$

Find the derivative of $s(t)=\sqrt{2t+1}$ .

Rewrite as $s(t)=(2t+1)^{1/2}$ and use the chain rule.

$s^{\prime}(t)=(2t+1)^{−1/2}$

Derivatives of Inverse Trigonometric Functions

The Main Idea

Key Derivatives:
$\begin{array}{lllll}\frac{d}{dx}(\sin^{-1} x)=\large \frac{1}{\sqrt{1-x^2}} & & & & \frac{d}{dx}(\cos^{-1} x)=\large \frac{-1}{\sqrt{1-x^2}} \\ \frac{d}{dx}(\tan^{-1} x)=\large \frac{1}{1+x^2} & & & & \frac{d}{dx}(\cot^{-1} x)=\large \frac{-1}{1+x^2} \\ \frac{d}{dx}(\sec^{-1} x)=\large \frac{1}{|x|\sqrt{x^2-1}} & & & & \frac{d}{dx}(\csc^{-1} x)=\large \frac{-1}{|x|\sqrt{x^2-1}} \end{array}$
For composite functions like $\sin^{-1}(g(x))$ , use the chain rule in conjunction with inverse trig derivatives
Pay attention to the domains of inverse trigonometric functions when differentiating

Use the inverse function theorem to find the derivative of $g(x)=\tan^{-1} x$ .

The inverse of $g(x)$ is $f(x)= \tan x$ .

$g^{\prime}(x)=\dfrac{1}{1+x^2}$

Find the derivative of $h(x)=x^2 \sin^{-1} x$

By applying the product rule, we have

h^{'} (x) = 2 x \sin^{- 1} x + \frac{1}{\sqrt{1 - x^{2}}} \cdot x^{2}

$h^{\prime}(x)=2x \sin^{-1} x+\frac{1}{\sqrt{1-x^2}} \cdot x^2$ .

Watch the following video to see the worked solution to this example.

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You can view the transcript for this segmented clip of “3.7 Derivatives of Inverse Functions (edited)” here (opens in new window).

Find the derivative of $h(x)= \cos^{-1} (3x-1)$

$h^{\prime}(x)=\frac{-3}{\sqrt{6x-9x^2}}$

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

$f^{\prime}(0)$ gives the slope of the tangent line.

$y=x$