Question

If $f(x)=\cos ^{-1}\left(2 x^2-1\right), x \in[-1,1]$. Then,

Accepted Answer

$f(x)$ is differentiable on $(-1,0) \cup(0,1)$ such that $f'(x) = \left\{ {\begin{array}{*{20}{l}} {\frac{{ - 2}}{{\sqrt {1 - {x^2}} }},}&{0 < x < 1}\ {\frac{2}{{\sqrt {1 - {x^2}} }},}&{ - 1 < x < 0} \end{array}} ight.$

Answer

$f(x)$ is differentiable on $(-1,1)$ such that $f^{\prime}(x)=\frac{-2}{\sqrt{1-x^2}}$

Answer

$f(x)$ is differentiable on $(-1,0) \cup(0,1)$ such that $f'(x) = \frac{{ - 2}}{{\sqrt {1 - {x^2}} }}$

Answer

$f(x)$ is differentiable on $(-1,1)$ such that $f'(x) = \left\{ {\begin{array}{*{20}{l}} {\frac{{ - 2}}{{\sqrt {1 - {x^2}} }},}&{0 \le x < 1}\ {\frac{2}{{\sqrt {1 - {x^2}} }},}&{ - 1 < x \le 0} \end{array}} ight.$

JEE MATHEMATICS MCQ: DIFFERENTIATION

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