d dx [ (1 - x^2 )^2 ]= - Vidyadip

MCQ

${d \over {dx}}[\cos {(1 - {x^2})^2}]$=

A
$ - 2x(1 - {x^2})\sin {(1 - {x^2})^2}$
B
$ - 4x(1 - {x^2})\sin {(1 - {x^2})^2}$
✓
$4x(1 - {x^2})\sin {(1 - {x^2})^2}$
D
$ - 2(1 - {x^2})\sin {(1 - {x^2})^2}$

✓

Answer

Correct option: C.

$4x(1 - {x^2})\sin {(1 - {x^2})^2}$

c
(c) $\frac{d}{{dx}}[\cos {(1 - {x^2})^2}] = - \sin {(1 - {x^2})^2}\frac{d}{{dx}}{(1 - {x^2})^2}$

$ = 4x(1 - {x^2})\sin {(1 - {x^2})^2}$.

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