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Question Number 9306 by tawakalitu last updated on 29/Nov/16

$$\mathrm{Differentiate}\mathrm{from}\mathrm{the}\mathrm{first}\mathrm{principle}:$$$$\mathrm{y}=\tan 2\mathrm{x}$$

Answered by mrW last updated on 30/Nov/16

$$\mathrm{y}\left(\mathrm{x}\right)=\tan \left(2\mathrm{x}\right)$$$$\mathrm{y}(\mathrm{x}+\mathrm{h})=\tan (2\mathrm{x}+2\mathrm{h})=\frac{\tan \left(2\mathrm{x}\right)+\tan \left(2\mathrm{h}\right)}{1-\tan \left(2\mathrm{x}\right)\times \tan \left(2\mathrm{h}\right)}$$$$\mathrm{y}(\mathrm{x}+\mathrm{h})-\mathrm{y}\left(\mathrm{x}\right)=\frac{\tan \left(2\mathrm{x}\right)+\tan \left(2\mathrm{h}\right)}{1-\tan \left(2\mathrm{x}\right)\times \tan \left(2\mathrm{h}\right)}-\tan \left(2\mathrm{x}\right)$$$$=\frac{\tan \left(2\mathrm{x}\right)+\tan \left(2\mathrm{h}\right)-\tan \left(2\mathrm{x}\right)+\tan {}^2\left(2\mathrm{x}\right)\times \tan \left(2\mathrm{h}\right)}{1-\tan \left(2\mathrm{x}\right)\times \tan \left(2\mathrm{h}\right)}$$$$=\frac{1+\tan {}^2\left(2\mathrm{x}\right)}{\frac{1}{\tan \left(2\mathrm{h}\right)}-\tan \left(2\mathrm{x}\right)}$$$$\frac{\mathrm{y}(\mathrm{x}+\mathrm{h})-\mathrm{y}\left(\mathrm{x}\right)}{\mathrm{h}}=\frac{1+\tan {}^2\left(2\mathrm{x}\right)}{\frac{\mathrm{h}}{\tan \left(2\mathrm{h}\right)}-\tan \left(2\mathrm{x}\right)\times \mathrm{h}}$$$$\underset{\mathrm{h}\to 0}{\lim }\frac{\mathrm{h}}{\tan \left(2\mathrm{h}\right)}=\underset{\mathrm{h}\to 0}{\lim }\frac{1}{2\times \frac{\tan \left(2\mathrm{h}\right)}{2\mathrm{h}}}=\frac{1}{2}$$$$\underset{\mathrm{h}\to 0}{\lim }\tan \left(2\mathrm{x}\right)\times \mathrm{h}=0$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\underset{\mathrm{h}\to 0}{\lim }\frac{\mathrm{y}(\mathrm{x}+\mathrm{h})-\mathrm{y}\left(\mathrm{x}\right)}{\mathrm{h}}=\frac{1+\tan {}^2\left(2\mathrm{x}\right)}{\frac{1}{2}}=2[1+\tan {}^2\left(2\mathrm{x}\right)]=\frac{2}{\cos {}^2\left(2\mathrm{x}\right)}$$

Commented by tawakalitu last updated on 29/Nov/16

$$\mathrm{Thanks}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}.$$

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