Chapter 3.1, Problem 12E

To determine

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

Answer to Problem 12E

The required value of $\frac{d}{dx}\left(3x^2\right)$ is 6x .

Explanation of Solution

Given information:

The function: $f\left(x\right)=3x^2$

Formula used:

Definition of the derivative:

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

Calculation:

Consider the given function be $f\left(x\right)=3x^2$ .

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

Write the above formula using the given function as:

  $\begin{array}{c}\frac{d}{dx}f\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{3{\left(x+h\right)}^2-3x^2}{h}\\ =\underset{h\to 0}{\lim }\frac{\cancel{3x^2}+6hx+3h^2-\cancel{3x^2}}{h}\\ =\underset{h\to 0}{\lim }\frac{6hx+3h^2}{h}\end{array}$

Simplify the limit further.

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{6hx+3h^2}{h}=\underset{h\to 0}{\lim }\frac{\cancel{h}\left(6x+3h\right)}{\cancel{h}}\\ =\underset{h\to 0}{\lim }\left(6x+3h\right)\end{array}$

Substitute 0 for h in the above limit and simplify.

  $\begin{array}{c}\underset{h\to 0}{\lim }\left(6x+3h\right)=6x+3\left(0\right)\\ =6x+0\\ =6x\end{array}$

Hence, the required value of $\frac{d}{dx}\left(3x^2\right)$ is 6x .