Match the functions in List-I with their deri | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The derivatives of the given inverse functions are as follows:$(A)$ $\frac{d}{dx}(\sec^{-1} x) = \frac{1}{|x|\sqrt{x^2-1}}$ for $|x| > 1$. This ma

Vedclass · Accepted Answer

$A-III, B-I, C-IV, D-II$

Vedclass · Answer

(D) The derivatives of the given inverse functions are as follows:$(A)$ $\frac{d}{dx}(\sec^{-1} x) = \frac{1}{|x|\sqrt{x^2-1}}$ for $|x| > 1$. This matches $III$.$(B)$ $\frac{d}{dx}(	anh^{-1} x) = \frac{1}{1-x^2}$ for $x \in (-1, 1)$. This matches $I$.$(C)$ $\frac{d}{dx}(\coth^{-1} x) = \frac{1}{1-x^2}$ for $x \in R - [-1, 1]$. This matches $IV$.$(D)$ $\frac{d}{dx}(\operatorname{cosech}^{-1} x) = \frac{-1}{|x|\sqrt{x^2+1}}$ for $x 
eq 0$. This matches $II$.Thus, the correct matching is $A-III, B-I, C-IV, D-II$.

List-$I$	List-$II$
$A$. $\sec^{-1} x$	$I$. $\frac{1}{1-x^2}, x \in (-1, 1)$
$B$. $\tanh^{-1} x$	$II$. $\frac{-1}{\|x\| \sqrt{x^2+1}}, x \neq 0$
$C$. $\coth^{-1} x$	$III$. $\frac{1}{\|x\| \sqrt{x^2-1}}, \|x\| > 1$
$D$. $\operatorname{cosech}^{-1} x$	$IV$. $\frac{1}{1-x^2}, x \in R - [-1, 1]$
	$V$. $\frac{-1}{\|x\| \sqrt{1-x^2}}, \|x\| < 1, x \neq 0$

Similar Questions

If $f(1)=1$ and $f^{\prime}(1)=3$,then the derivative of $f(f(f(x)))+(f(x))^2$ at $x=1$ is

If $y = f \left( \frac{3x + 4}{5x + 6} \right)$ and $f'(x) = \tan(x^2)$,then $\frac{dy}{dx} = $

Let $f(x) = (x^x)^x$ and $g(x) = x^{(x^x)}$,then:

If $y = \tan^{-1}(\frac{1}{x^2+x+1}) + \tan^{-1}(\frac{1}{x^2+3x+3}) + \tan^{-1}(\frac{1}{x^2+5x+7}) + \dots$ upto $n$ terms, then $y'(0) = $

$A$ function $f: R \rightarrow R$ is such that $y f(x+y) + \cos(mxy) = 1 + y f(x)$. If $m=2$, then $f'(x) =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module