Differentiate the following from the first principlex^2 - Vidyadip

Question

Differentiate the following from the first principle$\text{x}^2\sin\text{x}$

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Answer

We have, $\text{f}(\text{x})=\text{x}^2\sin\text{x}$
$\because\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{(\text{x}+\text{h})^2\sin(\text{x}+\text{h})-\text{x}^2\sin\text{x}}{\text{h}}$
$\lim_\limits{\text{h}\rightarrow0}\frac{(\text{x}^2+\text{h}^2+\text{2hx})(\sin\text{x}.\cos\text{x}+\cos\text{x}\sin\text{h})-\text{x}^2\sin\text{x}}{\text{h}}$ $[\because\sin(\text{A}+\text{B})=\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}]$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{x}^2\sin\text{x}(\cos\text{h}-1)}{\text{h}}+\frac{\text{h}(\text{h}+\text{2x})\sin\text{x}.\cos\text{h}}{\text{h}}+(\text{x}+\text{h})^2\cos\text{x}\frac{\sin\text{h}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}-\text{x}^2\sin\text{x}\times\frac{2\sin^2\frac{\text{h}}{2}}{\Big(\frac{\text{h}}{2}\Big)^2}\times\frac{\text{h}^2}{4}+(\text{h}+\text{2x})\sin\text{x}.\cos\text{2h}+(\text{x}+\text{h})^2\cos\text{x}$
$=0+(2\text{x}\sin\text{x}+\text{x}^2\cos\text{x})$
$=\text{2x}\sin\text{x}+\text{x}^2\cos\text{x}$

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