Question
Differentiate the following from first principle
$\tan\text{x}^2$
$\tan\text{x}^2$
$\text{f}(\text{x})=\tan\text{x}^2$
$\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\tan{(\text{x}+\text{h})^2}-\tan{\text{x}^2}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\frac{\sin(\text{x}+\text{h})^2}{\cos(\text{x}+\text{h})}-\frac{\sin\text{x}^2}{\cos\text{x}^2}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\frac{\sin(\text{x}+\text{h})^2\cos\text{x}^2-\cos(\text{x}+\text{h})^2\sin\text{x}^2}{\cos(\text{x}+\text{h})^2\cos\text{x}^2}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\Big((\text{x}+\text{h})^2-\text{x}^2\Big)}{\text{h}.\cos(\text{x}+\text{h})^2.\cos\text{x}^2}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{x}^2+\text{h}^2+\text{2xh}-\text{x}^2)}{\text{h}.\cos(\text{x}+\text{h})^2.\cos\text{x}^2}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{h}^2+\text{2xh})}{\text{h}.\cos(\text{x}+\text{h})^2.\cos\text{x}^2 }$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\text{h}}{\text{h}}\times\frac{(\text{h}+\text{2x})}{\cos(\text{x}+\text{h})^2.\cos\text{x}^2}$
$=1.\frac{\text{2x}}{\cos^2(\text{x})^2}$
$=\text{2x}\sec^2\text{x}^2$
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$\frac{\text{x}\sin\text{x}}{1+\cos\text{x}}$