MCQ
${d \over {dx}}(\sin 2{x^2})$ equals
  • $4x\cos \,(2{x^2})$
  • B
    $2\sin {x^2}\cos {x^2}$
  • C
    $4x\sin ({x^2})$
  • D
    $4x\sin ({x^2})\cos ({x^2})$

Answer

Correct option: A.
$4x\cos \,(2{x^2})$
a
(a) $\frac{d}{{dx}}(\sin 2{x^2}) = \cos 2{x^2}.\frac{d}{{dx}}2{x^2} = \cos 2{x^2}.4x = 4x.\cos 2{x^2}$.

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

Similar questions

If $A$ is a $m \times n$ matrix such that $A B$ and $B A$ are both defined, then $B$ is an
The corner points of the feasible region determined by the system of linear inequalities are (0, 0), (4, 0), (2, 4) and (0, 5). If the maximum value of z = ax + by, where a, b > 0 occurs at both (2, 4) and (4, 0), then:
The value of p and q for which the function $f ( x )=\left\{\begin{array}{cl}\frac{\sin (p+1) x+\sin x}{x} & , x<0 \\ \frac{q}{x^2} & , x=0 \\ \frac{\sqrt{x+b x^2}-\sqrt{x}}{x^{\frac{3}{2}}} & , x>0\end{array}\right.$ is continuous for all $x \in R$, are
The derivative of $\cos^{-1}(2\text{x}^2-1)$ with respect to $\cos^{-1}\text{x}$ is:
If $\int_{}^{} {(\cos x - \sin x)\;dx = \sqrt 2 \sin (x + \alpha ) + c} $, then $\alpha = $
The interval in which the function $f(x)=2 x^3+9 x^2+$ $12 x-1$ is decreasing, is
If the curve $ay +x^2=7$ and $x^3=y$ cut orthogonally at $(1,1)$, then a is equal to :
Choose the correct answer from the given four options:
The area of the region bounded by the curve $\text{y}=\sin\text{x}$ between the ordinates x = 0, $\text{x}=\frac{\pi}{2}$ and the x-axis is:
Choose the correct option from given four options : $\int\frac{\cos2\text{x}-\cos2\theta}{\cos\text{x}-\cos\theta}\text{dx}$ is equal to :
Let $P(x)$ be a polynomial with real coefficients such that $P\left(\sin ^2 x\right)=P\left(\cos ^2 x\right)$ for all $x \in[0, \pi / 2)$. Consider the following statements:

$I.$ $P(x)$ is an even function.

$II.$ $P(x)$ can be expressed as a polynomial in $(2 x-1)^2$

$III.$ $P(x)$ is a polynomial of even degree.

Then,