2.9 Derivatives of Inverse Trigonometric Functions
Derivatives of Inverse Trigonometric Functions
$
\begin{array}{ll}
\frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^2}} & \frac{d}{d x}\left(\cos ^{-1} x\right)=-\frac{1}{\sqrt{1-x^2}} \\
\frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^2} & \frac{d}{d x}\left(\cot ^{-1} x\right)=-\frac{1}{1+x^2} \\
\frac{d}{d x}\left(\sec ^{-1} x\right)=\frac{1}{x \sqrt{x^2-1}} & \frac{d}{d x}\left(\csc ^{-1} x\right)=-\frac{1}{x \sqrt{x^2-1}}
\end{array}
$
Example
- Differentiate $y=x \tan ^{-1} x$.
▶️Answer/Explanation
Solution
$
\begin{aligned}
& \text { प } \frac{d y}{d x}=\frac{d}{d x}\left(x \tan ^{-1} x\right) \\
& \quad=x \frac{d}{d x} \tan ^{-1} x+\tan ^{-1} x \frac{d}{d x}(x) \\
& \quad=x \cdot \frac{1}{1+x^2}+\tan ^{-1} x \cdot 1 \\
& \quad=\frac{x}{1+x^2}+\tan ^{-1} x
\end{aligned}
$
Example
- Differentiate $y=\frac{1}{\cos ^{-1} x}$.
▶️Answer/Explanation
Solution
$
\begin{aligned}
y & =\frac{1}{\cos ^{-1} x}=\left(\cos ^{-1} x\right)^{-1} \\
\frac{d y}{d x} & =\frac{d}{d x}\left(\cos ^{-1} x\right)^{-1} \\
& =-1\left(\cos ^{-1} x\right)^{-2} \frac{d}{d x}\left(\cos ^{-1} x\right) \\
& =-\frac{1}{\left(\cos ^{-1} x\right)^2} \cdot-\frac{1}{\sqrt{1-x^2}} \\
& =\frac{1}{\left(\cos ^{-1} x\right)^2 \sqrt{1-x^2}}
\end{aligned}
$
Power Chain Rule
Exercises – Derivatives of Inverse Trigonometric Functions
Multiple Choice Questions
Example
- 1. $\frac{d}{d x}\left(\arcsin x^2\right)=$
(A) $-\frac{2 x}{\sqrt{1-x^2}}$
(B) $\frac{2 x}{\sqrt{x^2-1}}$
(C) $\frac{2 x}{\sqrt{x^4-1}}$
(D) $\frac{2 x}{\sqrt{1-x^4}}$
▶️Answer/Explanation
Ans:D
Example
- 2. If $f(x)=\arctan \left(e^{-x}\right)$, then $f^{\prime}(-1)=$
(A) $\frac{-e}{1+e}$
(B) $\frac{e}{1+e}$
(C) $\frac{-e}{1+e^2}$
(D) $\frac{-1}{1+e^2}$
▶️Answer/Explanation
Ans: C
Example
- If $f(x)=\arctan (\sin x)$, then $f^{\prime}\left(\frac{\pi}{3}\right)=$
(A) $\frac{2}{7}$
(B) $\frac{1}{2}$
(C) $\frac{\sqrt{2}}{3}$
(D) $\frac{\sqrt{3}}{3}$
▶️Answer/Explanation
Ans:A
Example
- If $y=\cos \left(\sin ^{-1} x\right)$, then $y^{\prime}=$
(A) $-\frac{1}{\sqrt{1-x^2}}$
(B) $-\frac{x}{\sqrt{1-x^2}}$
(C) $\frac{2 x}{\sqrt{1-x^2}}$
(D) $-\frac{2 x}{\sqrt{x^2-1}}$
▶️Answer/Explanation
Ans:B