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Derivatives — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSDerivatives4 Marks
Question
Differentiate the following from first principle$\cos\sqrt{\text{x}}$
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Answer

We have,$\text{f}(\text{x})=\cos\sqrt{\text{x}}$
$\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sqrt{\cos(\text{x}+\text{h})}-\cos\sqrt{\text{x}}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{-2\sin\Big(\frac{\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}}}{2}\Big)\sin\Big(\frac{\sqrt{(\text{x}+\text{h})}+\sqrt{\text{x}}}{2}\Big)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{-2\sin\Bigg(\frac{\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}}}{2}\Bigg)(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})(\sqrt{(\text{x}+\text{h})}+\sqrt{\text{x}})}{\Bigg(\frac{(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})}{2}\Bigg)(\sqrt{(\text{x}+\text{h})}+\sqrt{\text{x}})\text{h}}\sin\Bigg(\frac{(\sqrt{(\text{x}+\text{h})}+\sqrt{\text{x}})}{2}\Bigg)$$=\lim_\limits{\text{h}\rightarrow0}\frac{-\sin\Bigg(\frac{(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})}{2}\Bigg)}{\Bigg(\frac{(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})}{2}\Bigg)}\times\frac{{\text{x}+\text{h}}-{\text{x}}}{(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})\text{h}}\times\sin\Bigg(\frac{\sqrt{(\text{x}+\text{h})+\text{x}}}{2}\Bigg)$
$=\lim_\limits{\text{h}\rightarrow0}-1\frac{\text{h}}{\text{h}(\sqrt{(\text{x}+\text{h})}+\sqrt{\text{x}})}\times\sin\Bigg(\frac{\sqrt{\text{x}+\text{h}}+\sqrt{\text{x}}}{2}\Bigg)$
$=\lim_\limits{\text{h}\rightarrow0}\frac{-1}{(\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}})}\sin\frac{\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}}}{2}$
$=\frac{-\sin\sqrt{\text{x}}}{2\sqrt{\text{x}}}$

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