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USING 1^ST PRINCIPLE, FIND dy/dx OF:

a. e^{ROOT OVER tanx}

b. tan root over x

$Let {f(x) = e}^{\sqrt{\tan x}} then {f(x+h)=e}^{\sqrt{\tan (x + h)}} by first principle of differenciation f^{1} (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{e^{\sqrt{\tan (x + h)}} - e^{\sqrt{\tan x}}}{h} = \lim_{h \to 0} e^{\sqrt{\tan x}} (\frac{e^{\sqrt{\tan (x + h)} - \sqrt{\tan x} - 1}}{h}) = e^{\sqrt{\tan x}} \lim_{h \to 0} {\frac{e^{\sqrt{\tan (x + h)} - \sqrt{\tan x} - 1}}{\sqrt{\tan (x + h)} - \sqrt{\tan x}} \times \frac{\sqrt{\tan (x + h)} - \sqrt{\tan x}}{h}} = e^{\sqrt{\tan x}} \lim_{h \to 0} {\frac{e^{\sqrt{\tan (x + h)} - \sqrt{\tan x} - 1}}{\sqrt{\tan (x + h)} - \sqrt{\tan x}}} \times \lim_{h \to 0} \frac{\sqrt{\tan (x + h)} - \sqrt{\tan x}}{h} = e^{\sqrt{\tan x}} \times 1 \times \lim_{h \to 0} \frac{\tan (x + h) - \tan x}{h} \times \frac{1}{\sqrt{\tan (x + h)} - \sqrt{\tan x}} = e^{\sqrt{\tan x}} \times \lim_{h \to 0} \frac{\sinh}{h (\cos (x + h) \cos x)} \times \frac{1}{\sqrt{\tan (x + h)} - \sqrt{\tan x}} = e^{\sqrt{\tan x}} \times \frac{1}{\cos^{2} x} \times \frac{1}{2 \sqrt{\tan x}} = \frac{e^{\sqrt{\tan x}} \sec^{2} x}{2 \sqrt{\tan x}}$

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