Inverse Trigonometric Functions

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