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? 
 A: 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.
A: Your work is not quite correct.
$$\begin{align}
(e^{k\tan\sqrt x})' &= e^{k\tan\sqrt x}\cdot(k\tan\sqrt x)' \\
 &= e^{k\tan\sqrt x}\cdot k\cdot (\tan\sqrt x)' \\
 &= e^{k\tan\sqrt x}\cdot k\cdot\sec^2\sqrt x\cdot (\sqrt x)' \\
 &= e^{k\tan\sqrt x}\cdot k\cdot\sec^2\sqrt x\cdot \frac 1{2\sqrt x} \\
 &= \frac{k\sec^2\sqrt x\cdot e^{k\tan\sqrt x}}{2\sqrt x}
\end{align}$$
A: Hint:
$y=e^{f(g(x))} $then$ \frac{dy}{dx}=e^{f(g(x))}f^{'}(g(x))g^{'}(x)$
In your case.    $ g(x)=\sqrt{x}$
$=>g^{'}(x)=\frac{1}{2\sqrt{x}}$
And $ f(x)=tan(\sqrt{x})$
Then $ f^{'}(x)=\sec^2(\sqrt{x})$
Now , 
$\frac{dy}{dx}=e^{k\tan(\sqrt{x})}\sec^2(\sqrt{x})\frac{k}{2\sqrt{x}}$
