Chapter 3.1, Problem 11E

To determine

To calculate: The value of $\frac{d}{dx}\left(x^2\right)$ .

Answer to Problem 11E

The required value of $\frac{d}{dx}\left(x^2\right)$ is 2x .

Explanation of Solution

Formula used:

Definition of the derivative:

  $\frac{dy}{dx}=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

Calculation:

Consider the given function be $y=x^2$ .

To find the value of $\frac{dy}{dx}$ , use the formula $\frac{dy}{dx}=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$ .

Write the above formula using the given function as:

  $\begin{array}{c}\frac{dy}{dx}=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{{\left(x+h\right)}^2-x^2}{h}\\ =\underset{h\to 0}{\lim }\frac{\cancel{x^2}+2hx+h^2-\cancel{x^2}}{h}\\ =\underset{h\to 0}{\lim }\frac{2hx+h^2}{h}\end{array}$

Simplify the limit further.

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{2hx+h^2}{h}=\underset{h\to 0}{\lim }\frac{\cancel{h}\left(2x+h\right)}{\cancel{h}}\\ =\underset{h\to 0}{\lim }\left(2x+h\right)\end{array}$

Substitute 0 for h in the above limit and simplify.

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

Hence, the required value of $\frac{d}{dx}\left(x^2\right)$ is 2x .