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Question Number 21464 by mondodotto@gmail.com last updated on 24/Sep/17

Answered by myintkhaing last updated on 24/Sep/17

$$\mathrm{y}+\delta \mathrm{y}=\sqrt{\tan (\mathrm{x}+\delta \mathrm{x})}$$$$\delta \mathrm{y}=\sqrt{\tan (\mathrm{x}+\delta \mathrm{x})}-\sqrt{\mathrm{tanx}}$$$$=(\sqrt{\tan (\mathrm{x}+\delta \mathrm{x})}-\sqrt{\mathrm{tanx}})\frac{\sqrt{\tan (\mathrm{x}+\delta \mathrm{x})}+\sqrt{\mathrm{tanx}}}{\sqrt{\tan (\mathrm{x}+\delta \mathrm{x})}+\sqrt{\mathrm{tanx}}}$$$$=\frac{\tan (\mathrm{x}+\delta \mathrm{x})-\mathrm{tanx}}{\sqrt{\tan (\mathrm{x}+\delta \mathrm{x})}+\sqrt{\mathrm{tanx}}}=\frac{\frac{\mathrm{tanx}+\tan \delta \mathrm{x}}{1-\mathrm{tanxtan}\delta \mathrm{x}}-\mathrm{tanx}}{\tan \sqrt{(\mathrm{x}+\delta \mathrm{x})}+\sqrt{\mathrm{tanx}}}$$$$=\frac{\tan \delta \mathrm{x}(1+{\tan }^2\mathrm{x})}{(1-\mathrm{tanxtan}\delta \mathrm{x})(\sqrt{\tan (\mathrm{x}+\delta \mathrm{x})}+\sqrt{\mathrm{tanx})}}$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\underset{\delta \mathrm{x}\to 0}{\lim }\frac{\tan \delta \mathrm{x}(1+{\tan }^2\mathrm{x})}{\delta \mathrm{x}(1-\mathrm{tanxtan}\delta \mathrm{x})(\sqrt{\tan (\mathrm{x}+\delta \mathrm{x})}+\sqrt{\mathrm{tanx}})}$$$$=\underset{\delta \mathrm{x}\to 0}{\lim }\frac{\tan \delta \mathrm{x}}{\delta \mathrm{x}}\underset{\delta \mathrm{x}\to 0}{\lim }\frac{{\sec }^2\mathrm{x}}{(1-\mathrm{tanxtan}\delta \mathrm{x})(\sqrt{\tan (\mathrm{x}+\delta \mathrm{x})}+\sqrt{\mathrm{tanx}})}$$$$=\left(1\right)\frac{{\sec }^2\mathrm{x}}{\left(1\right)2\sqrt{\mathrm{tanx}}}$$$$\text{You can't use 'macro parameter character \#' in math mode}$$

Answered by $@ty@m last updated on 24/Sep/17

$$\frac{dy}{dx}=\underset{\delta x\to 0}{\lim }\frac{\sqrt{\tan (x+\delta x)}-\sqrt{\tan x}}{\delta x}$$$$\frac{dy}{dx}=\underset{\delta x\to 0}{\lim }\frac{\sqrt{\tan (x+\delta x)}-\sqrt{\tan x}}{\tan (x+\delta x)-\tan x}\times \frac{\tan (x+\delta x)-\tan x}{\delta x}$$$$=L_1\times L_2$$$$where{L}_1=\underset{\delta x\to 0}{\lim }\frac{\sqrt{\tan (x+\delta x)}-\sqrt{\tan x}}{\tan (x+\delta x)-\tan x}$$$$=\frac{1}{2\sqrt{\tan x}},usingformula\underset{x\to a}{\lim }\frac{x^n-a^n}{x-a}={na}^{n-1p}$$$$and{L}_2=\underset{\delta x\to 0}{\lim }\frac{\tan (x+\delta x)-\tan x}{\delta x}$$$$=\underset{\delta x\to 0}{\lim }\frac{1}{\delta x}[\frac{\sin (x+\delta x)}{\cos (x+\delta x)}-\frac{\sin x}{\cos x}]$$$$=....$$$$=...$$$$={\sec }^{2}x\left\{Doityourself\right\}$$$$\therefore \frac{dy}{dx}=\frac{{sec}^{2}x}{2\sqrt{\tan x}}$$

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