Putting g ( x ) = e x {\displaystyle g(x)=e^{x}} and h ( x ) = 3 x {\displaystyle h(x)={\sqrt {3x}}} , we can write the given function f {\displaystyle f} as a composition function
f ( x ) = e 3 x = g ( h ( x ) ) {\displaystyle f(x)=e^{\sqrt {3x}}=g(h(x))} .
Then, by the chain rule, we have the derivative of f {\displaystyle f} ;
f ′ ( x ) = g ′ ( h ( x ) ) ⋅ h ′ ( x ) = e 3 x ( 3 x ) ′ {\displaystyle f'(x)=g'(h(x))\cdot h'(x)=e^{\sqrt {3x}}({\sqrt {3x}})'} .
To get the derivative of 3 x {\displaystyle {\sqrt {3x}}} , we again use the chain rule with y = 3 x {\displaystyle y={\sqrt {3x}}} and u = 3 x {\displaystyle u=3x} ;
( 3 x ) ′ = d y d x = d y d u ⋅ d u d x = d d u u ⋅ d d x 3 x = 1 2 u ⋅ 3 = 3 2 3 x {\displaystyle ({\sqrt {3x}})'={\frac {dy}{dx}}={\frac {dy}{du}}\cdot {\frac {du}{dx}}={\frac {d}{du}}{\sqrt {u}}\cdot {\frac {d}{dx}}3x={\frac {1}{2{\sqrt {u}}}}\cdot 3={\frac {3}{2{\sqrt {3x}}}}} .
Plugging this back to the equation, we get
f ′ ( x ) = e 3 x 3 2 3 x {\displaystyle \color {blue}f'(x)=e^{\sqrt {3x}}{\frac {3}{2{\sqrt {3x}}}}} .
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, we again use the chain rule with 
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