Chapter 12, Problem 67RE

To determine

Find the Derivative of a function.

Answer to Problem 67RE

  $g\text{'}\left(x\right)=4x$

Explanation of Solution

Given information:

Find the derivative of the function.

  $g\left(x\right)=2x^2-1$

Calculation:

Consider the function,

  $g\left(x\right)=2x^2-1$

Derivative of the function $g$ at any point $x$ is,

  $g\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{g\left(x+h\right)-g\left(x\right)}{h}$ Provided the limit exists.

Determine the limit,

  $\begin{array}{l}g\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{g\left(x+h\right)-g\left(x\right)}{h}\\ \underset{h\to 0}{\lim }\frac{g\left(x+h\right)-g\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{\left\{2{\left(x+h\right)}^2-1\right\}-\left(2x^2-1\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{2\left(x^2+2xh+h^2\right)-2x^2}{h}\\ =\underset{h\to 0}{\lim }\frac{2\left(2xh+h^2\right)}{h}\end{array}$

The determine expression is in the form of $\frac{0}{0}$

  $\underset{h\to 0}{\lim }\frac{2\left(2xh+h^2\right)}{h}=\underset{h\to 0}{\lim }\frac{2h\left(2x+h\right)}{h}$

Since $h\to 0$ so $h\ne 0$

Now cancelled $h$ we get,

  $\begin{array}{l}\underset{h\to 0}{\lim }\frac{2h\left(2x+h\right)}{h}\\ =\underset{h\to 0}{\lim }2\left(2x+h\right)\\ =2\left(2x+0\right)\\ =4x\end{array}$

Hence the derivative of the function $g$ is $g\text{'}\left(x\right)=4x$