If $f(x) = \cos^{-1} x$,$g(x) = e^x$,and $h(x) = g(f(x))$,then $\frac{h'(x)}{h(x)} = $

If $f(1)=1$ and $f^{\prime}(1)=3$,then the derivative of $f(f(f(x)))+(f(x))^2$ at $x=1$ is

If $y=\frac{\cos x}{1+\sin x}$,then
$(a)$ $\frac{dy}{dx}=\frac{-1}{1+\sin x}$
$(b)$ $\frac{dy}{dx}=\frac{1}{1+\sin x}$
$(c)$ $\frac{dy}{dx}=-\frac{1}{2} \sec ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)$
$(d)$ $\frac{dy}{dx}=\frac{1}{2} \sec ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)$

The solution set of $f'(x) > g'(x)$,where $f(x) = \frac{1}{2} (5^{2x+1})$ and $g(x) = 5^x + 4x \ln 5$ is:

Let $f(x)$ and $g(x)$ be twice differentiable functions satisfying $f''(x) = g''(x)$ for all $x \in R$, $f'(1) = 2g'(1) = 4$ and $g(2) = 3f(2) = 9$. Then $f(25) - g(25)$ is equal to:

If $y = \log \tan \sqrt{x}$,then the value of $\frac{dy}{dx}$ is