Differentiate the following from the first principlex - Vidyadip

Question

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

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Answer

We have,$\text{f}(\text{x})=\text{x}\sin\text{x}$
$\because\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{(\text{x}+\text{h})\sin(\text{x}+\text{h})-\text{x}\sin\text{x}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{x}(\sin(\text{x}+\text{h})-\sin\text{x})}{\text{h}}+\sin(\text{x}+\text{h})\ $ $\Big[\sin\text{c}-\sin\text{d}=2\cos\frac{\text{c}+\text{d}}{2}\sin\frac{\text{c}-\text{d}}{2}\Big]$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{x}\times2\cos\Big(\text{x}+\frac{\text{h}}{2}\Big)\sin\frac{\text{h}}{2}}{\text{h}}+\sin(\text{x}+\text{h})\ $ $\Big[\because\lim_\limits{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$
$=\text{2x}\times\cos\text{x}\times\frac{1}{2}+\sin\text{x}$
$=\text{x}\times\cos\text{x}+\sin\text{x}$
$=\sin\text{x}+\text{x}\cos\text{x}$

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