If $f(x)=\int\limits_{1/x}^{\sqrt{x}}cos\,t^2dt(x>0)$ then $\frac{df(x)}{dx}$ is
Answer & explanation
Correct answer: option 2
$\frac{df(x)}{dx}=cos(\sqrt{x})^2\frac{d\sqrt{x}}{dx}-cos(\frac{1}{x})^2.\frac{d}{dx}(\frac{1}{x})$
$=\frac{1}{2\sqrt{x}}cos\,x+\frac{cos\,x^{-2}}{x^2}=\frac{x\sqrt{x}cos\,x+2cos(x^{-2})}{2x^2}$
Hence (B) is the correct answer.