Answer: [tex]f'(x)=\dfrac{9\cos(9\ln (x))}{x}[/tex].
Step-by-step explanation:
The given function is
[tex]f(x)=\sin(9\ln (x))[/tex]
Using chain rule differentiate w.r.t. x.
[tex]f'(x)=\cos(9\ln (x))\dfrac{d}{dx}(9\ln (x))[/tex] [tex]\left[\because \dfrac{d}{dx}\sin x=\cos x\right][/tex]
[tex]f'(x)=\cos(9\ln (x))\left[9\dfrac{d}{dx}(\ln (x))\right][/tex]
[tex]f'(x)=\cos(9\ln (x))\left[9\times \dfrac{1}{x}\right][/tex] [tex]\left[\because \dfrac{d}{dx}\ln x=\dfrac{1}{x}\right][/tex]
[tex]f'(x)=\dfrac{9\cos(9\ln (x))}{x}[/tex]
Therefore, [tex]f'(x)=\dfrac{9\cos(9\ln (x))}{x}[/tex].
