Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Inverse Trigonometric Functions

Question:

Derivative of $\sin(\cos(x^2))$ with respect to $x^2 $ is :

Options:

$-2xcos (sin(x^2))$

$-2xcos (cos(x^2))sin(x^2)$

$-cos (cos(x^2))sin(x^2)$

$cos (cos(x^2))sin(x^2)$

Correct Answer:

$-cos (cos(x^2))sin(x^2)$

Explanation:

The correct answer is Option (3) → $-\cos (\cos(x^2))\sin(x^2)$

$y=\sin(\cos(x^2))$,  $z=x^2$

$\frac{dy}{dx}=\cos(\cos(x^2))(-\sin(x^2))2x$

$\frac{dz}{dx}=2x$

$⇒\frac{dy}{dz}=-\cos (\cos(x^2))\sin(x^2)$