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Derivatives — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsDerivatives4 Marks
Question
Differentiate the following from the first principle$\text{f}(\text{x})=\frac{\sin\text{x}}{\text{x}}$
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Answer

We have, $\text{f}(\text{x})=\frac{\sin\text{x}}{\text{x}}$
$\because\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\frac{\sin(\text{x}+\text{h})}{\text{x}+\text{h}}-\frac{\sin\text{x}}{\text{x}}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{x}\sin(\text{x}+\text{h})-(\text{x}+\text{h})\sin\text{x}}{\text{xh}(\text{x}+\text{h})}$
$\lim_\limits{\text{h}\rightarrow0}\frac{\text{x}(\sin\text{x}.\cos\text{h}+\cos\text{h}.\sin\text{h})-\text{x}\sin\text{x}-\text{h}.\sin\text{x}}{\text{xh}(\text{x}+\text{h})}$ $[\because\sin(\text{A}+\text{B}=\sin\text{A}.\cos{B}+\cos\text{A}.\sin\text{B})]$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{x}.\sin\text{x}(\cos\text{h}-1)}{\text{xh}(\text{x}+\text{h})}+\frac{\text{x}\cos\text{x}.\sin\text{x}}{\text{xh}(\text{x}+\text{h})}-\frac{\text{h}\sin\text{x}}{\text{xh}(\text{x}+\text{h})}$ $\Big[\because1-\cos\text{h}=2\sin^2\frac{\text{h}}{2}\Big]$
$=\frac{-\text{x}\sin\text{x}}{\text{x}(\text{x}+\text{h})}\times\frac{2\sin^2\frac{\text{h}}{2}}{\frac{\text{h}^2}{4}}\times\frac{\text{h}}{4}+\frac{\text{x}\cos\text{x}}{\text{x}^2}-\frac{\sin\text{x}}{\text{x}^2}$
$\because\text{h}\rightarrow0\Rightarrow\frac{\text{h}}{2}\rightarrow0$ and $\lim_\limits{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1$
$=0+\frac{\text{x}\cos\text{x}-\sin\text{x}}{\text{x}^2}$
$=\frac{\text{x}\cos\text{x}-\sin\text{x}}{\text{x}^2}$

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