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Derivatives — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsDerivatives5 Marks
Question
Differentiate the following from the first principle$\cos\Big(\text{x}-\frac{\pi}{8}\Big)$
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Answer

Let $\text{f}(\text{x})=\cos\Big(\text{x}-\frac{\pi}{8}\Big).$ Then, $\text{f}(\text{x}+\text{h})=\cos\Big(\text{x}+\text{h}-\frac{\pi}{8}\Big)$$\therefore\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\lim_\limits{\text{h}\rightarrow0}\frac{\cos\Big(\text{x}+\text{h}-\frac{\pi}{8}\Big)-\cos\Big(\text{x}-\frac{\pi}{8}\Big)}{\text{h}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\lim_\limits{\text{h}\rightarrow0}\frac{-2\sin\Bigg[\frac{\big(\text{x}+\text{h}-\frac{\pi}{8}\big)+\big(\text{x}-\frac{\pi}{8}\big)}{2}\Bigg]\sin\Bigg[\frac{\big(\text{x}+\text{h}-\frac{\pi}{8}\big)-\big(\text{x}-\frac{\pi}{8}\big)}{2}\Bigg]}{\text{h}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\lim_\limits{\text{h}\rightarrow0}\frac{-2\sin\Bigg[\frac{\text{2x}+\text{h}-\frac{2\pi}{8}}{2}\Bigg]\sin\Big[\frac{\text{h}}{2}\Big]}{\text{h}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\lim_\limits{\text{h}\rightarrow0}-\sin\Bigg[\frac{2\text{x}+\text{h}-\frac{2\pi}{8}}{2}\Bigg]\times\lim_\limits{\text{h}\rightarrow0}\frac{\sin\Big[\frac{\text{h}}{2}\Big]}{\frac{\text{h}}{2}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=-\sin\Bigg[\frac{\text{2x}+0-\frac{2\pi}{8}}{2}\Bigg]\times1$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=-\sin\Big(\text{x}-\frac{\pi}{8}\Big)$

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