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Calculus I Limits and Derivatives

Sunil Karna

Section 3.3 Derivatives of Inverse Functions

If a function \(f(x)\) is differentiable and has an inverse function \(f^{-1}(x)\text{,}\) then the derivative of the inverse function at a point is the reciprocal of the derivative of the original function at the corresponding point.
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Let
\begin{equation*} y = f^{-1}(x)\text{,} \end{equation*}
then
\begin{equation*} x = f(y) \end{equation*}
Now differentiate on both side w.r.t. \(x\) using chain rule.
\begin{equation*} 1= f'(y) = \frac{dy}{dx} \end{equation*}
Then:
\begin{equation*} \frac{dy}{dx} = \frac{1}{f'(y)} \end{equation*}
but, \(y = f^{-1}(x)\text{,}\) thus,
\begin{equation*} y' = (f^{-1})'(x)= \frac{1}{f'(f^{-1}(x))} \end{equation*}
provided \(f'(f^{-1}(x)) \neq 0. \)
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Example 3.3.1.

Let
\begin{equation*} y = e^x\text{,} \end{equation*}
Now, take logarithm on both sides, we get
\begin{equation*} \ln y = \ln e^{x} = x\ln e = x \end{equation*}
Replacing \(x\) and \(y\text{,}\) we get the corresponding point:
\begin{equation*} y = \ln x \end{equation*}
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Derivatives of Inverse Functions
  1. \begin{equation*} \frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}} \end{equation*}
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  2. \begin{equation*} \frac{d}{dx}(\cos^{-1} x) = -\frac{1}{\sqrt{1-x^2}} \end{equation*}
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  3. \begin{equation*} \frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2} \end{equation*}
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  4. \begin{equation*} \frac{d}{dx}(\cot^{-1} x) = -\frac{1}{1+x^2} \end{equation*}
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  5. \begin{equation*} \frac{d}{dx}(\sec^{-1} x) = \frac{1}{|x|\sqrt{x^2-1}} \end{equation*}
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  6. \begin{equation*} \frac{d}{dx}(\csc^{-1} x) = -\frac{1}{|x|\sqrt{x^2-1}} \end{equation*}
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Example 3.3.2.

Find the derivative of \(\sin^{-1} x \) using first principle.
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Solution.
If
\begin{equation*} f(x) = \sin^{-1} x \end{equation*}
\begin{equation*} \text{ but } \, f'(x) = \lim_{h \to 0} \frac{f(x+h) -f(x)}{h} \end{equation*}
\begin{equation*} \therefore \, f'(\sin^{-1} x) = \lim_{h \to 0} \frac{\sin^{-1}(x+h) -\sin^{-1}(x)}{h} \end{equation*}
Let
\begin{equation*} \sin^{-1}(x+h) = \alpha \, \text{ and } \sin^{-1}(x) = \beta \end{equation*}
\begin{equation*} \text{ then }\, \sin \alpha = (x+h) \, \text {and } \, \sin \beta = x \end{equation*}
\begin{equation*} \text{ also }\, h = \sin \alpha - \sin \beta \end{equation*}
\begin{equation*} \text{ as }\, h \to 0 = \sin \alpha \to \sin \beta \, \implies \, \alpha \to \beta \quad \text {or } \, (\alpha - \beta) \to 0 \end{equation*}
\begin{equation*} \text{Now } \, f'(\sin^{-1} x) = \lim_{(\alpha - \beta) \to 0} \frac{\alpha -\beta}{ \sin \alpha - \sin \beta} \end{equation*}
\begin{equation*} = \frac{1}{\lim_{(\alpha - \beta) \to 0} \left[\frac{ \sin \alpha - \sin \beta}{\alpha -\beta} \right]} \end{equation*}
\begin{equation*} = \frac{1}{\lim_{(\alpha - \beta) \to 0} \left[\frac{ 2 \cos \left(\frac{\alpha + \beta}{2}\right) \sin \left(\frac{\alpha - \beta}{2}\right)}{\frac{\alpha -\beta}{2}\times 2} \right]} \end{equation*}
\begin{equation*} = \frac{1}{\lim_{(\alpha \to \beta)} \cos \left(\frac{\alpha + \beta}{2}\right)} \times \frac{1}{\lim_{(\alpha-\beta) \to 0 } \frac{\sin \left(\frac{\alpha - \beta}{2}\right)}{\frac{\alpha - \beta}{2}}} \end{equation*}
\begin{equation*} =\frac{1}{\cos \beta} = \frac{1}{\sqrt{1-\sin^2 \beta}} = \frac{1}{\sqrt{1-x^2}} \end{equation*}
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Example 3.3.3.

Find the derivative of \(\cos^{-1} x \) using first principle.
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Solution.
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Example 3.3.4.

Find the derivative of \(\tan^{-1} x \) using first principle.
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Solution.
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Example 3.3.5.

Find the derivative of \(\sin^{-1} x.\)
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Solution.
Let
\begin{equation*} y = \sin^{-1} x \end{equation*}
so
\begin{equation*} x=\sin y \, \text{ where } -\frac{\pi}{2}\leq y \leq \frac{\pi}{2} \end{equation*}
Differentiate w.r.t "x", we get-
\begin{equation*} 1= \cos y \frac{dy}{dx} \end{equation*}
\begin{equation*} or, \, \frac{dy}{dx} = \frac{1}{\cos y } = \frac{1}{\cos (\sin^{-1} x) } \end{equation*}
To simplify, within the given boundary, \(-\frac{\pi}{2}\leq y \leq \frac{\pi}{2}\) consider a right angled triangle with hypoteneous equal to \(1\) and perpendicular equals \(x.\) as shown in figure below.
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\begin{equation*} \cos (\sin^{-1} x) = \cos y =\sqrt{1-x^2} \end{equation*}
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Example 3.3.6.

Find \(\frac{dy}{dx}\) if
  1. \begin{equation*} y=\sin^{-1}(2x\sqrt{1-x^2}); \, -\frac{1}{\sqrt{2}} \lt x \lt \frac{1}{\sqrt{2}}. \end{equation*}
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  2. \begin{equation*} y=\sin^{-1}(2x\sqrt{1-x^2}); \, \frac{1}{\sqrt{2}} \lt x \lt 1. \end{equation*}
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  3. \begin{equation*} y=\sin^{-1}(3x-4x^3); \, \frac{1}{\sqrt{2}} \lt x \lt 1. \end{equation*}
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  4. \begin{equation*} y=\tan^{-1}\left(\sqrt{\frac{1-\cos 2x}{1+\cos 2x}}\right); \, -\frac{\pi}{2} \lt x \lt \frac{\pi}{2}. \end{equation*}
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  5. \begin{equation*} y=\tan^{-1}\left(\frac{1-\cos x}{\sin x}\right); \, -\pi \lt x \lt \pi. \end{equation*}
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  6. \begin{equation*} y=\tan^{-1}\left(\frac{a\cos x-b\sin x}{b\cos x+a\sin x}\right) \, \end{equation*}
    where \(-\frac{\pi}{2}\lt x \lt \frac{\pi}{2}\) and \(\frac{a}{b}\tan x \gt -1.\)
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Example 3.3.7.

Let \(g(x)\) be the inverse of \(f(x).\) If \(f'(x)=\frac{x^5}{1-x^2}, \) and \(f(a)=3\) then find \(g'(3).\)
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Solution.
\begin{align*} g(x) \amp =f^{-1} (x) \\ f(g(x)) \amp = x \end{align*}
diff. w.r.t. "x"
\begin{equation*} f'(g(x))g'(x) = 1 \end{equation*}
\begin{equation*} g'(x) = \frac{1}{f'(g(x))} \end{equation*}
At \(x=3,\)
\begin{equation*} g'(3) = \frac{1}{f'(g(3))} = \frac{1}{f'(a)} \,\text{ put } a = g(3) \end{equation*}
From
\begin{equation*} f'(x)=\frac{x^5}{1-x^2} \end{equation*}
\begin{equation*} f'(a)=\frac{a^5}{1-a^2} \end{equation*}
\begin{equation*} \therefore \, g'(3)=\frac{1-a^2}{a^5} \end{equation*}
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