d dx ^ - 1 (3x - 4 x^3 ) = - Vidyadip

MCQ

${d \over {dx}}{\sin ^{ - 1}}(3x - 4{x^3}) = $

✓
${3 \over {\sqrt {1 - {x^2}} }}$
B
${{ - 3} \over {\sqrt {1 - {x^2}} }}$
C
${1 \over {\sqrt {1 - {x^2}} }}$
D
${{ - 1} \over {\sqrt {1 - {x^2}} }}$

✓

Answer

Correct option: A.

${3 \over {\sqrt {1 - {x^2}} }}$

a
(a) Put $x = \sin \theta ,$ we get $\frac{d}{{dx}}{\sin ^{ - 1}}(3x - 4{x^3})$

$ = \frac{d}{{dx}}{\sin ^{ - 1}}(\sin 3\theta ) = \frac{3}{{\sqrt {1 - {x^2}} }}$.

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