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LIMITS AND DERIVATIVES — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSLIMITS AND DERIVATIVES4 Marks
Question
Find the derivative of cos x from first principle.
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Answer

Let $ \text{f}(\text{x})=\cos\text{x}$. accordingly, from the first principle, $\text{f}'(\text{x})=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\cos(\text{x}+\text{h})-\cos\text{x}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\bigg[\frac{\cos\text{x}\cos\text{h}-\sin\text{x}\sin\text{h}-\cos\text{x}}{\text{h}}\bigg]$ $=\lim\limits_{\text{h}\rightarrow0}\bigg[\frac{-\cos\text{x}(1-\cos\text{h})-\sin\text{x}\sin\text{h}}{\text{h}}\bigg]$ $=\lim\limits_{\text{h}\rightarrow0}\bigg[\frac{-\cos\text{x}(1-\cos\text{h})}{\text{h}}-\frac{\sin\text{x}\sin\text{h}}{\text{h}}\bigg]$ $=-\cos\text{x}\bigg(\lim\limits_{\text{h}\rightarrow0}\frac{1-\cos\text{h}}{\text{h}}\bigg)-\sin\text{x}\lim\limits_{\text{h}\rightarrow0}\bigg(\frac{\sin\text{h}}{\text{h}}\bigg)$ $=-\cos\text{x}(0)-\sin\text{x}(1)$ $\bigg[\lim\limits_{\text{h}\rightarrow0}\frac{1-\cos\text{h}}{\text{h}}=0\ \text{and} \ \lim\limits_{\text{h}\rightarrow0}\frac{\sin\text{h}}{\text{h}}=1\bigg]$ $=-\sin\text{x}$ $\therefore\text{f}'(\text{x})=-\sin\text{x}$

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