Answer:
The answer is "[tex]\frac{5 e^{5\sqrt{x} }}{2\sqrt{x}}[/tex]".
Step-by-step explanation:
Given:
[tex]y = e^{{\color{\red}5}\sqrt{x}}[/tex]
let
[tex]\to t= 5\sqrt{x}\\\\\frac{dt}{dx}= 5 \frac{1}{2\sqrt{x}}\\\\\frac{dt}{dx}= \frac{5}{2\sqrt{x}}\\\\[/tex]
and
[tex]\to y=e^t\\\\\to \frac{dy}{dt}=e^t\\[/tex]
[tex]\to \frac{dy}{dt}=e^{5\sqrt{x} }\\[/tex]
So,
[tex]\to \frac{dy}{dx}= \frac{dy}{dt} \times \frac{dt}{dx}[/tex]
[tex]=e^{5\sqrt{x} }\times \frac{5}{2\sqrt{x}}\\\\= \frac{5 e^{5\sqrt{x} }}{2\sqrt{x}}[/tex]
OR
[tex]\to g(x) = 5\sqrt{x} \\\\\to f(x) = e^{(x)}\\\\[/tex]
Derivate:
[tex]\to f''g' = \frac{e^{(5\sqrt{x})}5}{(2\sqrt{x})}[/tex]