Differentiate the Function: $y=e^{k\ tan\sqrt{x}}$ $y=e^{k\tan\sqrt{x}}$
$=e^{k\tan\sqrt{x}}\cdot [{k\tan\sqrt{x}}]'$
$=e^{k\tan\sqrt{x}}\cdot\ (k)\cdot[\tan x^{\frac{1}{2}}]'+(\tan x^{\frac{1}{2}})\cdot[k]'$
$=e^{k\tan\sqrt{x}}\cdot\ (k)\cdot (\frac{1}{2}tanx)\cdot \sec^2x + (\tan x^{\frac{1}{2}})\cdot (1)$
Is my process correct? 

An alternative solution: $$y = e^{k \tan \sqrt{x}} \implies \ln y = k \tan \sqrt{x},$$so: $$\frac{y'}{y} = \frac{k \sec^2\sqrt{x}}{2\sqrt{x}} \implies y' = e^{k \tan \sqrt{x}}\frac{k \sec^2\sqrt{x}}{2\sqrt{x}}.$$

I used the following facts:


*

*chain rule on $\ln y$ to get $y'/y$ on the left side.

*the derivative of $\sqrt{x}$ is $1/(2\sqrt{x})$.

*the derivative of $\tan u$ is $\sec^2 u$.

*the chain rule on $\tan(\sqrt{x})$ along with the two last items.
