If $f(x)=e^x$ and $h(x)=(f \circ f)(x)$,then $\frac{h^{\prime}(x)}{h(x)}=$

The relation $R$ is defined as $R = \{(4, 5), (1, 4), (4, 6), (7, 6), (3, 7)\}$. Then $R^{-1} o R$ is:

Let the functions $f:(-1,1) \rightarrow R$ and $g:(-1,1) \rightarrow(-1,1)$ be defined by $f(x)=|2 x-1|+|2 x+1|$ and $g(x)=x-[x]$,where $[x]$ denotes the greatest integer less than or equal to $x$. Let $f \circ g:(-1,1) \rightarrow R$ be the composite function defined by $(f \circ g)(x)=f(g(x))$. Suppose $c$ is the number of points in the interval $(-1,1)$ at which $f \circ g$ is $NOT$ continuous,and suppose $d$ is the number of points in the interval $(-1,1)$ at which $f \circ g$ is $NOT$ differentiable. Then the value of $c+d$ is.

If $f(x) = \frac{(\tan 1^{\circ}) x + \log_{e}(123)}{x \log_{e}(1234) - (\tan 1^{\circ})}$,$x > 0$,then the least value of $f(f(x)) + f(f(4/x))$ is $...........$.

If $f(x)=e^{|x|}$ and $g(x)=\log x$,then $(g \circ f)(x) =$

Let $R$ be the set of real numbers and the mapping $f: R \rightarrow R$ and $g: R \rightarrow R$ be defined by $f(x) = 5 - x^2$ and $g(x) = 3x - 4$,then the value of $(f \circ g)(-1)$ is