Q. 48 - 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.     

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

Step-by-Step Solution

Answer

The derivative of the function is $\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{3}{2}{x}^2-7x\end{array}$.

Step 1. Given Information

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

Step 2. Simplify the function

Simplify the given function.

$\begin{array}{rcl}f\left(x\right)& =& \frac{x^4}{2x}-\frac{7x^3}{2x}\\ & =& \frac{x^3}{2}-\frac{7}{2}{x}^2\end{array}$

Step 3. Find the derivative

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

$\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{d}{dx}\left(\frac{x^3}{2}\right)-\frac{d}{dx}(\frac{7}{2}{x}^2)\end{array}$

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

$f\text{'}\left(x\right)=\frac{1}{2}\begin{array}{rcl}& & \frac{d}{dx}\left(x^3\right)-\frac{7}{2}\frac{d}{dx}(x^2)\end{array}$

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

$\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{1}{2}\left(3x^2\right)-\frac{7}{2}(2x)\\ & =& \frac{3}{2}{x}^2-7x\end{array}$