What is the derivative of e^tanx by first principle?

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Solution

$Let y = e^{\tan x} \Rightarrow y + ∆ y = e^{\tan (x + ∆ x)} \Rightarrow ∆ y = e^{\tan (x + ∆ x)} - e^{\tan x} \Rightarrow \frac{∆ y}{∆ x} = \frac{e^{\tan (x + ∆ x)} - e^{\tan x}}{∆ x} \Rightarrow \frac{∆ y}{∆ x} = e^{\tan x} [\frac{e^{\tan (x + ∆ x) - \tan x} - 1}{∆ x}] \Rightarrow \lim_{∆ x \to 0} \frac{∆ y}{∆ x} = e^{\tan x} \lim_{∆ x \to 0} [\frac{e^{\tan (x + ∆ x) - \tan x} - 1}{\tan (x + ∆ x) - \tan x}] \times \frac{\tan (x + ∆ x) - \tan x}{∆ x} \Rightarrow \frac{dy}{dx} = e^{\tan x} \times 1 \times \lim_{∆ x \to 0} \frac{\frac{\sin (x + ∆ x)}{\cos (x + ∆ x)} - \frac{\sin x}{\cos x}}{∆ x} \Rightarrow \frac{dy}{dx} = e^{\tan x} \times \lim_{∆ x \to 0} \frac{\sin (x + ∆ x) . \cos x - \cos (x + ∆ x) . \sin x}{∆ x} \times \frac{1}{\cos x . \cos (x + ∆ x)} \Rightarrow \frac{dy}{dx} = e^{\tan x} \times \lim_{∆ x \to 0} (\frac{\sin ∆ x}{∆ x}) \times \lim_{∆ x \to 0} \frac{1}{\cos x . \cos (x + ∆ x)} \Rightarrow \frac{dy}{dx} = e^{\tan x} \times 1 \times \frac{1}{\cos^{2} x} = e^{\tan x} . \sec^{2} x$

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