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Question: If \(f(x) = \int_{x^{2}}^{x^{4}}{\sin\sqrt{t}dt,}\) then \(f^{'}(x)\) equals...

If f(x)=∫x2x4sin⁡tdt,f(x) = \int_{x^{2}}^{x^{4}}{\sin\sqrt{t}dt,} then f′(x)f^{'}(x) equals

A

sin⁡x2−sin⁡x\sin x^{2} - \sin x

B

4x3sin⁡x2−2xsin⁡x4x^{3}\sin x^{2} - 2x\sin x

C

x4sin⁡x2−xsin⁡xx^{4}\sin x^{2} - x\sin x

D

None of these

Answer

4x3sin⁡x2−2xsin⁡x4x^{3}\sin x^{2} - 2x\sin x

Explanation

Solution

We have f(x)=∫x2x4sin⁡tdtf(x) = \int_{x^{2}}^{x^{4}}{\sin\sqrt{t}}dt

∴f′(x)=ddx(x4)(sin⁡x4)−ddx(x2)(sin⁡x2)f'(x) = \frac{d}{dx}(x^{4})(\sin\sqrt{x^{4}}) - \frac{d}{dx}(x^{2})(\sin\sqrt{x^{2}})

=4x3sin⁡x2−2xsin⁡x= 4x^{3}\sin x^{2} - 2x\sin x.