Differentiate the following from the first principle (x+1) - Vidyadip

Question

Differentiate the following from the first principle

$\sin(\text{x}+1)$

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Answer

Let $\text{f}(\text{x})=\sin(\text{x}+1).$ Then, $\text{f}(\text{x}+1)=\sin\big((\text{x}+\text{h})+1\big)$

$\therefore\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$

$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\big((\text{x}+\text{h})+1\big)-\sin(\text{x}+1)}{\text{h}}$

$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\lim_\limits{\text{h}\rightarrow0}\frac{2\sin\Bigg[\frac{\big((\text{x}+\text{h})+1\big)-(\text{x}+1)}{2}\Bigg]\cos\Bigg[\frac{\big((\text{x}+\text{h})+1\big)+(\text{x}+1)}{2}\Bigg]}{\text{h}}$

$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\lim_\limits{\text{h}\rightarrow0}\frac{2\sin\Big[\frac{\text{h}}{2}\Big]\cos\Big[\frac{2\text{x}+2+\text{h}}{2}\Big]}{\text{h}}$

$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\Big[\frac{\text{h}}{2}\Big]}{\frac{\text{h}}{2}}\times\lim_\limits{\text{h}\rightarrow0}\cos\Big[\frac{2\text{x}+2+\text{h}}{2}\Big]$

$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=1\times\cos\Big[\frac{\text{2x}+2+0}{2}\Big]$

$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\cos(\text{x}+1)$

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