Differentiate the following from first principle x

Question

Differentiate the following from first principle$\tan\sqrt{\text{x}}$

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Answer

We have,$\text{f}(\text{x})=\tan\sqrt{\text{x}}$
$\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\tan\sqrt{(\text{x}+\text{h})}-\tan\sqrt{\text{x}}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}}}{\text{h}.\cos\sqrt{\text{x}+\text{h}}\cos\sqrt{\text{x}}} \ \Bigg[\because\tan\text{A}-\tan\text{B}=\frac{\sin(\text{A}-\text{B})}{\cos\text{A}.\cos\text{B}}\Bigg]$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})}{(\text{x}+\text{h}-\text{x})\cos\sqrt{\text{x}}.\cos\sqrt{\text{x}+\text{h}}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})}{(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})(\sqrt{(\text{x}+\text{h})}+\sqrt{\text{x}})\cos\sqrt{\text{x}}.\cos\sqrt{\text{x}+\text{h}}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})}{(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})}\times\frac{1}{(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})\cos\sqrt{\text{x}}.\cos\sqrt{\text{x}+\text{h}}}$
$=1\times\frac{1}{2\sqrt{\text{x}}\cos\sqrt{\text{x}}\cos\sqrt{\text{x}+\text{h}}}\ \Bigg[\because\lim_\limits{\text{h}\rightarrow0}=\frac{\sin(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})}{(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})}=1\Bigg]$
$=\frac{1}{2\sqrt{\text{x}}\cos^2\text{x}}$
$=\frac{\sec^2\text{x}}{2\sqrt{\text{x}}}$