[tex]f(x) = 3^x + \ln x -108 x\\\\f'(x) = 3^x \ln 3 + \dfrac 1x - 108\\\\f''(x) = 3^x \cdot 0+ \ln 3\cdot 3^x\cdot \ln (3)-x^{-1-1}-0= 3^x \cdot \ln^2 3 -x^{-2}\\ \\f'''(x) = 3^x \cdot 0 + \ln^2 (3) \cdot 3^x \cdot \ln (3) +2x^{-2-1} = 3^x \ln^3 (3)+2x^{-3}\\\\f''''(x) = 3^x \cdot 0 + \ln^3 (3) \cdot 3^x \ln(3) -3\cdot 2 x^{-3-1} = 3^x\ln^4 (3)-6x^{-4} \\\\\text{Hence the fourth derivative of f(x) is}~~ 3^x\ln^4 (3)-6x^{-4}.\\\\[/tex]
[tex]\textbf{Note:}\\\\\dfrac{d}{dx}(uv) = u \dfrac{dv}{dx} + v\dfrac{du}{dx}\\\\\dfrac{d}{dx} \ln (x) = \dfrac 1x\\\\\dfrac{d}{dx} a^x = a^x \ln a\\\\\dfrac{d}{dx} x^n = nx^{n-1}\\\\\dfrac{d}{dx} (\text{Constant}) = 0[/tex]
