Differentiate the following from the first principlex - Vidyadip

Question

Differentiate the following from the first principle$\text{x}\cos\text{x}$

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Answer

we have, $\text{f}(\text{x})=\text{x}\cos\text{x}$
$\because\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{(\text{x}+\text{h})\cos(\text{x}+\text{h})-\text{x}\cos\text{x}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{x}\cos(\text{x}+\text{h})\text{h}\cos(\text{x}+\text{h})-\text{x}\cos\text{x}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{x}\Big\{\cos(\text{x}+\text{h})-\cos\text{x}\Big\}}{\text{h}}+\cos(\text{x}+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{x}.2\sin\Big(\text{x}-\text{x}+\frac{\text{h}}{2}\Big)\sin\Big(\text{x}+\frac{\text{h}}{2}\Big)+\cos(\text{x}+\text{h})$ $\Big[\because\cos\text{}A-\cos\text{B}=2\sin\frac{\text{B}-\text{A}}{2}\sin\frac{\text{B}+\text{A}}{2}\Big]$
$=\lim_\limits{\text{h}\rightarrow0}2\text{x}.\sin\Big[\frac{-\text{h}}{2}\Big]\sin\Big(\text{x}+\frac{\text{h}}{2}\Big)+\cos(\text{x}+\text{h})$
$=-\text{x}\sin\text{x}+\cos\text{x}$

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