Suggested languages for you:

Americas

Europe

Q. 88

Expert-verifiedFound in: Page 199

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

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

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

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

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

Using the definition of derivative we get,

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

Hence proved

94% of StudySmarter users get better grades.

Sign up for free