The derivative of $sin^3(cos\, x^2)$ wit

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Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Inverse Trigonometric Functions

Question:

The derivative of $sin^3(cos\, x^2)$ with respect to x i s:

Options:

$-6x\sin^2(\cos x^2)\cos(\cos x^2)\sin^2x$

$-6x\, sin^2(cos\, x^2)sin\, x$

$-6x\, sin^3(cos\, x^2)sin\, x^2$

$-6x\, sin^3(cos\, x^2)$

Correct Answer:

$-6x\sin^2(\cos x^2)\cos(\cos x^2)\sin^2x$

Explanation:

The correct answer is option (1) → $-6x\sin^2(\cos x^2)\cos(\cos x^2)\sin^2x$ $\frac{80}{89}$

$\frac{d}{dx}\sin^3(\cos x^2)$

$=3\sin^2(\cos x^2)\cos(\cos x^2)(-\sin x^2)(2x)$

$=-6x\sin^2(\cos x^2)\cos(\cos x^2)\sin^2x$