The function y = sin^{-1}(cos x) is not diffe | vedclass

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

Question

(D) The function $y = \sin^{-1}(\cos x)$ is defined when $-1 \le \cos x \le 1$. However,the derivative of $\sin^{-1} u$ is $\frac{1}{\sqrt{1-u^2}}$,wh

Vedclass · Accepted Answer

All options are correct

Vedclass · Answer

(D) The function $y = \sin^{-1}(\cos x)$ is defined when $-1 \le \cos x \le 1$. However,the derivative of $\sin^{-1} u$ is $\frac{1}{\sqrt{1-u^2}}$,which is undefined when $u = \pm 1$.Thus,the function $y = \sin^{-1}(\cos x)$ is not differentiable where $\cos x = 1$ or $\cos x = -1$.For $\cos x = -1$,$x = (2n+1)\pi$ for any integer $n$. For $n=0$,$x = \pi$.For $\cos x = 1$,$x = 2n\pi$ for any integer $n$. For $n=1$,$x = 2\pi$,and for $n=-1$,$x = -2\pi$.Since the function is not differentiable at $x = \pi$,$x = 2\pi$,and $x = -2\pi$,all the given options are correct.

The function $y = \sin^{-1}(\cos x)$ is not differentiable at . . . . . .

Similar Questions

Given $f(x) = \begin{cases} 1 + x & x < 0 \\ 2 - 3x & x \geq 0 \end{cases}$,find the critical point $x = \dots \dots$.

If the function $f(x) = \begin{cases} \frac{(e^{kx}-1) \sin kx}{4 \tan x}, & x \neq 0 \\ P, & x=0 \end{cases}$ is differentiable at $x=0$, then

Prove that the function $f$ given by $f(x) = |x - 1|, x \in R$ is not differentiable at $x = 1$.

Let $S$ be the set of points where the function $f(x) = |2 - |x - 3||, x \in R,$ is not differentiable. Then $\sum_{x \in S} f(f(x))$ is equal to

The function which is continuous for all real values of $x$ and differentiable at $x = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module