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Expert-verified Found in: Page 199 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # Use the definition of the derivative to prove the following special case of the product rule$\frac{d}{dx}\left({x}^{2}f\left(x\right)\right)=2xf\left(x\right)+{x}^{2}f\text{'}\left(x\right)$

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

See the step by step solution

## Step 1: Given information

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

## Step 2: Find the derivative

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

Using the definition of derivative we get,

$\underset{h\to 0}{\mathrm{lim}}\frac{g\left(x+h\right)-g\left(x\right)}{h}\phantom{\rule{0ex}{0ex}}\underset{h\to 0}{\mathrm{lim}}\frac{{\left(x+h\right)}^{2}f\left(x+h\right)-{x}^{2}f\left(x\right)}{h}\phantom{\rule{0ex}{0ex}}\underset{h\to 0}{\mathrm{lim}}\frac{\left({x}^{2}+2xh+{h}^{2}\right)f\left(x+h\right)-{x}^{2}f\left(x\right)}{h}\phantom{\rule{0ex}{0ex}}\underset{h\to 0}{\mathrm{lim}}\frac{{x}^{2}\left(f\left(x+h\right)-f\left(x\right)\right)}{h}+\underset{h\to 0}{\mathrm{lim}}\frac{h\left(2x+h\right)f\left(x+h\right)}{h}\phantom{\rule{0ex}{0ex}}={x}^{2}f\text{'}\left(x\right)+2xf\left(x\right)$

Hence proved ### Want to see more solutions like these? 