Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Continuity and Differentiability

Question:

What is the derivative of $f(x)=2\sqrt{\cos(x^2)}$

Options:

$\frac{-2\sin(x^2)}{\cot(x^2)}$

$\frac{-2x\sin(x^2)}{\sqrt{\cos(x^2)}}$

$\tan^2x$

$\sin^2x$

Correct Answer:

$\frac{-2x\sin(x^2)}{\sqrt{\cos(x^2)}}$

Explanation:

$f(x)=2\sqrt{\cos(x^2)}$

so $f'(x)=\frac{-2x\sin(x^2)}{\sqrt{\cos(x^2)}}$