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If $f(x)=\int\limits_{1/x}^{\sqrt{x}}cos\,t^2dt(x>0)$ then $\frac{df(x)}{dx}$ is |
$\frac{\sqrt{x}cos\,x+2cos(x^{-2})}{2x\sqrt{x}}$ $\frac{x\sqrt{x}cos\,x+2cos(x^{-2})}{2x^2}$ $2\sqrt{x}cos\,x-\frac{2}{x}cos(\frac{1}{x})$ none of these |
$\frac{x\sqrt{x}cos\,x+2cos(x^{-2})}{2x^2}$ |
$\frac{df(x)}{dx}=cos(\sqrt{x})^2\frac{d\sqrt{x}}{dx}-cos(\frac{1}{x})^2.\frac{d}{dx}(\frac{1}{x})$ $=\frac{1}{2\sqrt{x}}cos\,x+\frac{cos\,x^{-2}}{x^2}=\frac{x\sqrt{x}cos\,x+2cos(x^{-2})}{2x^2}$ Hence (B) is the correct answer. |
