Question

Find the derivatives of the function:$f\left(x\right)=\frac{1}{x^4-5x^3+2}.$

Short answer

The derivative of$f\left(x\right)=\frac{1}{x^4-5x^3+2}is-\frac{4x^3-15x^2}{{(x^4-5x^3+2)}^2}$.

Step-by-step solution

Given information 

The given expression is$f\left(x\right)=\frac{1}{x^4-5x^3+2}.$

Simplification 

$$\begin{gathered} f\text{'}\left(x\right) \\ =\frac{d}{dx}\left(f\right(x\left)\right) \\ =\frac{d}{dx}\left(\frac{1}{x^4-5x^3+2}\right) \\ =-\frac{1}{{(x^4-5x^3+2)}^2}.\frac{d}{dx}(x^4-5x^3+2) \\ =-\frac{4x^3-15x^2}{{(x^4-5x^3+2)}^2} \\ =\frac{15x^2-4x^3}{{(x^4-5x^3+2)}^2} \\ Hence,f\text{'}\left(x\right)=\frac{15x^2-4x^3}{{(x^4-5x^3+2)}^2}. \end{gathered}$$