Answer:
[tex]f^{\prime}(x)=3(\sin x+x\cos x)e^{3x\sin x}[/tex]Explanation:
Given the function
[tex]f(x)=e^{3x\sin x}[/tex]f'(x) represents the first derivative of f(x) with respect to x.
This can be done by using the principle of function of a function.
Let u = 3x sinx
Then
f'(x) = f'(u).u'(x)
[tex]\begin{gathered} f(u)=e^u \\ f^{\prime}(u)=e^u \end{gathered}[/tex][tex]\begin{gathered} u(x)=3x\sin x \\ u^{\prime}(x)=3\sin x+3x\cos x \end{gathered}[/tex]Therefore;
[tex]f^{\prime}(x)=e^u.(3\sin x+3x\cos x)[/tex]with u = 3x sinx, we have
[tex]\begin{gathered} f^{\prime}(x)=(3\sin x+3x\cos x)e^{3x\sin x} \\ \\ =3(\sin x+x\cos x)e^{3x\sin x} \end{gathered}[/tex]