Q. 88 - Calculus

Question

Use the definition of the derivative to prove the following special case of the product rule

$\frac{d}{dx}\left(x^{2}f\right(x\left)\right)=2xf\left(x\right)+x^{2}f\text{'}\left(x\right)$

Step-by-Step Solution

Answer

We proved the special case of product function using the definition of the derivative

Step 1: Given information

We are given a function $\frac{d}{dx}\left(x^{2}f\right(x\left)\right)=2xf\left(x\right)+x^{2}f\text{'}\left(x\right)$

Step 2: Find the derivative

Consider $g\left(x\right)=x^{2}f\left(x\right)$

Using the definition of derivative we get,

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

Hence proved