Differentiating an expression with f(x) in it $\frac{d}{dx}$ $\frac{x^4}{f(x)}$ at $x=1$ where $f(1)=1$ and $f'(1)=3$
I've tried simply differentiating the expression using the quotient rule, obtaining $\frac{4x^3*f(x)-x^4*f'(x)}{f(x)^2}$ But I'm not sure about where to go from here, I'm confused by the use of f(x) in the expression

Since $f$ is differentiable at $1$, the composition $x\mapsto f(x)/x^4$ is differentiable at $1$ aswell. Hence by the quotient rule
$$\left .\frac{\mathrm d}{\mathrm d x }\frac{f(x)}{x^4}\right|_{x=1} = \left.\frac{4x^3 f(x)-x^4 f'(x)}{f(x)^2}\right|_{x=1}=\frac{4-3}{1}=1,$$
where we used $f(1)=1$ and $f'(1)=3$.
