# Calculate the derivatives of the functions. f(x)=(4x-1)^{-1}

Derivatives
Calculate the derivatives of the functions.
$$\displaystyle{f{{\left({x}\right)}}}={\left({4}{x}-{1}\right)}^{{-{1}}}$$

2021-06-21

$$\displaystyle{f{{\left({x}\right)}}}={4}{x}-{1}{)}^{{-{1}}}$$ Differentiale both sides with respect to x
$$\displaystyle{f}'{\left({x}\right)}={\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{\left({4}{x}-{1}\right)}^{{-{1}}}\right]}$$
Apply the Generalized Power Rule $$\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{u}^{{{n}}}\right]}'=\nu^{{{n}-{1}}}{\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}}$$,

let $$u=4x-1$$
$$\displaystyle{f}'{\left({x}\right)}=-{\left({4}{x}-{1}\right)}^{{-{1}-{1}}}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{4}{x}-{1}\right]}$$
Therefore,
$$\displaystyle{f}'{\left({x}\right)}=-{\left({4}{x}-{1}\right)}^{{-{2}}}{\left({4}\right)}$$
Simplify
$$\displaystyle{f}'{\left({x}\right)}=-{\frac{{{4}}}{{{\left({4}{x}-{1}\right)}^{{{2}}}}}}$$