Question
Differentiate the following from first principle
$\tan^2\text{x} $
$\tan^2\text{x} $
$\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\tan^2(\text{x}+\text{h})-\tan^2\text{x}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\big\{\tan(\text{x}+\text{h})+\tan{\text{x}}\big\}\big\{\tan(\text{x}+\text{h})-\tan{\text{x}}\big\}}{\text{h}}$
$\big[\because\tan^2\text{A}-\tan^2\text{B}=(\tan\text{A}+\tan\text{B})(\tan\text{A}-\tan\text{B})\big]$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\frac{\sin(\text{x}+\text{h}+\text{x})}{\cos(\text{x}+\text{h})\cos\text{x}}\times\frac{\sin(\text{x}+\text{h}-\text{x})}{\cos(\text{x}+\text{h})\cos\text{x}}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{2x}+\text{h})}{\text{h}.\cos(\text{x}+\text{h})\cos\text{x}}\times\frac{\sin\text{h}}{\cos(\text{x}+\text{h})\cos\text{x}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\text{h}}{\text{h}}\times\frac{\sin2\text{x}}{\cos^2\text{x}.\cos^2(\text{x}+\text{h})}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin2\text{x}}{\cos^2\text{x}.\cos^2\text{x}}\ \Big[\because\lim_\limits{\text{h}\rightarrow0}\frac{\sin\text{h}}{\text{h}}=1\Big]$
$=\lim_\limits{\text{h}\rightarrow0}\frac{2\sin\text{x}.\cos\text{x}}{\cos^2\text{x}}\times\frac{1}{\cos^2\text{x}}\ [\sin2\text{x}=2\sin\text{x}\cos\text{x}]$
$=\lim_\limits{\text{h}\rightarrow0}2\tan\text{x}.\sec^2\text{x}$
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| | C1 | | C2 |
| (a) | Boys and girls alternate. | (i) | 5! × 6! |
| (b) | No two girls sit together. | (ii) | 10! – 5! 6! |
| (c) | All the girls sit together. | (iii) | (5!)2 + (5!)2 |
| (d) | All the girls are never together. | (iv) | 2! 5! 5! |
$\cos\sqrt{\text{x}}$