Solveeit Logo
Login

Question

Question: If f (x) = \(\int_{1/x}^{\sqrt{x}}{\cos t^{2}}\)dt (x \> 0) then \(\frac{df(x)}{dx}\) is...

If f (x) = ∫1/xxcos⁡t2\int_{1/x}^{\sqrt{x}}{\cos t^{2}}dt (x > 0) then df(x)dx\frac{df(x)}{dx} is

A

xcosx+2cos(x−2)2xx\frac{\sqrt{x}cosx + 2cos(x^{- 2})}{2x\sqrt{x}}

B

xxcosx+2cos(x−2)2x2\frac{x\sqrt{x}cosx + 2cos(x^{- 2})}{2x^{2}}

C

2xcosx−2xcos⁡(1x)2\sqrt{x}cosx - \frac{2}{x}\cos\left( \frac{1}{x} \right)

D

None of these.

Answer

xxcosx+2cos(x−2)2x2\frac{x\sqrt{x}cosx + 2cos(x^{- 2})}{2x^{2}}

Explanation

Solution

df(x)dx\frac{df(x)}{dx} = cos (x)2dxdx−cos⁡(1x)2.ddx(1x)\left( \sqrt{x} \right)^{2}\frac{d\sqrt{x}}{dx} - \cos{}\left( \frac{1}{x} \right)^{2}.\frac{d}{dx}\left( \frac{1}{x} \right)

= 12xcos⁡x+cos⁡x−2x2\frac{1}{2\sqrt{x}}\cos{}x + \frac{\cos x^{- 2}}{x^{2}}= xxcos⁡x+2cos⁡(x−2)2x2\frac{x\sqrt{x}\cos x + 2\cos(x^{- 2})}{2x^{2}}