Q. 47 - 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.    

$f\left(x\right)=\frac{x^2-1}{x+1}$

Step-by-Step Solution

Answer

The derivative of the function is $f\text{'}\left(x\right)=1$ such that $x\ne -1$.

Step 1. Given Information

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

Step 2. Simplify the function

• Apply the identity $a^2-b^2=(a-b)(a+b)$ in the numerator of the given function.

$\begin{array}{rcl}f\left(x\right)& =& \frac{(x-1)(x+1)}{x+1}\\ & =& x-1\end{array}$

• So, the simplified function is $f\left(x\right)=x-1$ such that $x\ne -1$.

Step 3. Find the derivative

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

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

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

$\begin{array}{rcl}f\text{'}\left(x\right)& =& 1-0\\ & =& 1\end{array}$