If $y=\sqrt{\cosh x+\sqrt{\cosh x}}$, then $\frac{d y}{d x}=$
$\frac{\sinh x\left(2 y^2+2 \cosh x+1\right)}{4 y\left(y^2+\cosh x\right)}$
$\frac{\sinh x\left(2 y^2-2 \cosh x-1\right)}{4 y\left(y^2-\cosh x\right)}$
$\frac{\sinh x(1-2 \sqrt{\cosh x})}{4 y \sqrt{\cosh x}}$
$\frac{\sinh x(1+2 \sqrt{\cosh x})}{4 y \sqrt{\cosh x})}$
$\frac{1}{x \sqrt{x^2-1}}$
$\frac{x+1}{x \sqrt{x^2-1}}$
$\frac{x+1}{x^2 \sqrt{x^2-1}}$
$\frac{x}{\sqrt{x^2-1}}$
If $y=(\log x)^{1 / x}+x^{\log x}$, at $x=e, \frac{d y}{d x}=$
$2+\frac{1}{e}$
$e^2+\frac{1}{2}$
$\frac{1}{e^2}+2$
$e+\frac{1}{e}$
The interval in which the function $f(x)=\tan ^{-1}(\sin x+\cos x)$ is an increasing function is
$\left(0, \frac{\pi}{2}\right)$
$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
$\left(-\frac{3 \pi}{4}, \frac{\pi}{4}\right)$
$\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$
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