If $y(\cos x)^{\sin x}=(\sin x)^{\sin x}$, then the value of $\frac{dy}{dx}$ at $x=\frac{\pi}{4}$ is

If $y=\sqrt{\frac{1-\sin ^{-1}(x)}{1+\sin ^{-1}(x)}}$,then $\frac{dy}{dx}$ at $x=0$ and $y=1$ is

If $u, v$ and $w$ are functions of $x,$ then show that $\frac{d}{d x}(u \cdot v \cdot w) = \frac{d u}{d x} \cdot v \cdot w + u \cdot \frac{d v}{d x} \cdot w + u \cdot v \cdot \frac{d w}{d x}$ in two ways - first by repeated application of product rule,second by logarithmic differentiation.

If $y=x^{\sqrt{x}}$, then $\frac{dy}{dx}=$

If $\frac{d}{d x} \left[ \frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x} \right] = f(x) \left[ \frac{2}{x+1} + \frac{1}{2(x-1)} - \frac{3}{x+4} - 1 \right]$,then $f(5) = $

If $y = {\left( {1 + \frac{1}{x}} \right)^x}$,then $\frac{dy}{dx} = $