Question
Differentiate the following from first principle
$\tan\sqrt{\text{x}}$
$\tan\sqrt{\text{x}}$
$\text{f}(\text{x})=\tan\sqrt{\text{x}}$
$\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\tan\sqrt{(\text{x}+\text{h})}-\tan\sqrt{\text{x}}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}}}{\text{h}.\cos\sqrt{\text{x}+\text{h}}\cos\sqrt{\text{x}}} \ \Bigg[\because\tan\text{A}-\tan\text{B}=\frac{\sin(\text{A}-\text{B})}{\cos\text{A}.\cos\text{B}}\Bigg]$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})}{(\text{x}+\text{h}-\text{x})\cos\sqrt{\text{x}}.\cos\sqrt{\text{x}+\text{h}}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})}{(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})(\sqrt{(\text{x}+\text{h})}+\sqrt{\text{x}})\cos\sqrt{\text{x}}.\cos\sqrt{\text{x}+\text{h}}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})}{(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})}\times\frac{1}{(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})\cos\sqrt{\text{x}}.\cos\sqrt{\text{x}+\text{h}}}$
$=1\times\frac{1}{2\sqrt{\text{x}}\cos\sqrt{\text{x}}\cos\sqrt{\text{x}+\text{h}}}\ \Bigg[\because\lim_\limits{\text{h}\rightarrow0}=\frac{\sin(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})}{(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})}=1\Bigg]$
$=\frac{1}{2\sqrt{\text{x}}\cos^2\text{x}}$
$=\frac{\sec^2\text{x}}{2\sqrt{\text{x}}}$
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| Column I | Column II | ||
| (a) | $1^2+2^2+3^2+....+\text{n}^2$ | (i) | $\Big[\frac{\text{n}(\text{n}+1)}{2}\Big]^2$ |
| (b) | $1^3+2^3+3^3+....\text{n}^3$ | (ii) | $\text{n}(\text{n}+1)$ |
| (c) | $2+4+6+....+2\text{n}$ | (iii) | $\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}$ |
| (d) | $1+2+3+....\text{n}$ | (iv) | $\frac{\text{n}(\text{n}+1)}{2}$ |