Answer
$f^{\prime}(x)=\dfrac{3^x x \ln 3-3^x \ln (3) -3^x}{(x-1)^2}$
Work Step by Step
We have: $f(x)=\dfrac{3^x}{x-1}$
We differentiate both sides with respect to $x$.
$f^{\prime}(x)=\dfrac{d}{dx} [\dfrac{3^x}{x-1}] \\=\dfrac{(3^x \ln 3)(x-1)-3^x}{(x-1)^2}$
Simplify to obtain:
$f^{\prime}(x)=\dfrac{3^x x \ln 3-3^x \ln (3) -3^x}{(x-1)^2}$
