Differentiate-from-the-first-principle-y-tan2x-

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

Differentiate from the first principle :

y = \tan 2 x

Answered by mrW last updated on 30/Nov/16

y (x) = \tan (2 x)

y (x + h) = \tan (2 x + 2 h) = \frac{\tan (2 x) + \tan (2 h)}{1 - \tan (2 x) \times \tan (2 h)}

y (x + h) - y (x) = \frac{\tan (2 x) + \tan (2 h)}{1 - \tan (2 x) \times \tan (2 h)} - \tan (2 x)

= \frac{\tan (2 x) + \tan (2 h) - \tan (2 x) + \tan^{2} (2 x) \times \tan (2 h)}{1 - \tan (2 x) \times \tan (2 h)}

= \frac{1 + \tan^{2} (2 x)}{\frac{1}{\tan (2 h)} - \tan (2 x)}

\frac{y (x + h) - y (x)}{h} = \frac{1 + \tan^{2} (2 x)}{\frac{h}{\tan (2 h)} - \tan (2 x) \times h}

\lim_{h \to 0} \frac{h}{\tan (2 h)} = \lim_{h \to 0} \frac{1}{2 \times \frac{\tan (2 h)}{2 h}} = \frac{1}{2}

\lim_{h \to 0} \tan (2 x) \times h = 0

\frac{dy}{dx} = \lim_{h \to 0} \frac{y (x + h) - y (x)}{h} = \frac{1 + \tan^{2} (2 x)}{\frac{1}{2}} = 2 [1 + \tan^{2} (2 x)] = \frac{2}{\cos^{2} (2 x)}

Commented by tawakalitu last updated on 29/Nov/16

Thanks sir . God bless you .