Q. 53 - Calculus

Question

Use the differentiation rules developed in this section to find the derivatives of the functions in Exercises 35-64. Note that it may be necessary to do some preliminary algebra before differentiating.    

$f\left(x\right)=\frac{\sqrt[5]{x^7}-2x^4}{x^3}$

Step-by-Step Solution

Answer

The derivative of the function is $\begin{array}{rcl}& & \frac{-8}{5}{x}^{-\frac{13}{5}}-2\end{array}$.

Step 1. Given Information

The given function is $f\left(x\right)=\frac{\sqrt[5]{x^7}-2x^4}{x^3}$.

Step 2. Simplify the function

Simplify the function.

$\begin{array}{rcl}f\left(x\right)& =& \frac{\sqrt[5]{x^7}}{x^3}-\frac{2x^4}{x^3}\\ & =& \frac{x^{{\displaystyle \frac{7}{5}}}}{x^3}-2x\\ & =& x^{\frac{7}{5}-3}-2x\\ & =& x^{-\frac{8}{5}}-2x\end{array}$

Step 3. Find the derivative

• Apply the difference rule of derivative, $(f-g)\text{'}\left(x\right)=f\text{'}\left(x\right)-g\text{'}\left(x\right)$.

$f\text{'}\left(x\right)=\frac{d}{dx}{x}^{-\frac{8}{5}}-\frac{d}{dx}\left(2x\right)$

• Apply constant multiple of derivative, $\left(kf\right)\text{'}\left(x\right)=kf\text{'}\left(x\right)$.

$f\text{'}\left(x\right)=\frac{d}{dx}{x}^{\frac{-8}{5}}-2\frac{d}{dx}\left(x\right)$

• Apply power rule of derivative, $\left(x^n\right)\text{'}=n{x}^{n-1}$.

$\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{-8}{5}\left(x^{\frac{-8}{5}-1}\right)-2\left(1\right)\\ & =& \frac{-8}{5}{x}^{-\frac{13}{5}}-2\end{array}$