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}$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

If $\cos \alpha + \cos \beta + \cos \gamma = \sin \alpha + \sin \beta + \sin \gamma = 0$ then $\cos 3\alpha + \cos 3\beta + \cos 3\gamma $ equals to

The sum to infinite term of the series $1 + \frac{2}{3} + \frac{6}{{{3^2}}} + \frac{{10}}{{{3^3}}} + \frac{{14}}{{{3^4}}} + \ldots \;$ is

The real values of $x$ and $y$ for which the equation $({x^4} + 2xi) - (3{x^2} + yi) = $$(3 - 5i) + (1 + 2yi)$ is satisfied, are

Matrix $\left[ {\begin{array}{*{20}{c}}0&{ - 4}&1\\4&0&{ - 5}\\{ - 1}&5&0\end{array}} \right]$is

A point moves in the $x-y$ plane such that the sum of its distances from two mutually perpendicular lines is always equal to $3$. The area enclosed by the locus of the point is- ............... $\mathrm{unit}^{2}$

Let $M=\left[\begin{array}{cc}\sin ^4 \theta & -1-\sin ^2 \theta \\ 1+\cos ^2 \theta & \cos ^4 \theta\end{array}\right]=\alpha I +\beta M ^{-1}$, where $\alpha=\alpha(\theta)$ and $\beta=\beta(\theta)$ are real number, and $I$ is the $2 \times 2$ identity matrix. If $\alpha^*$ is the minimum of the set $\{\alpha(\theta): \theta \in[0,2 \pi)\}$ and $\beta^*$ is the minimum of the set $\{\beta(\theta): \theta \in[0,2 \pi)\}$, then the value of $\alpha^*+\beta^*$ is

Let a circle tangent to the directrix of a parabola ${y^2} = 2ax$ has its centre coinciding with the focus of the parabola. Then the point of intersection of the parabola and circle is

If $x + \frac{1}{x} = \sqrt 3 ,$ then $x =$

If $y = (1 + \tan A)(1 - \tan B)$ where $A - B = \frac{\pi }{4}$, then ${(y + 1)^{y + 1}}$ is equal to

$\int_{}^{} {\frac{{3\sin x + 2\cos x}}{{3\cos x + 2\sin x}}\;dx = } $