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Difference \(e\sqrt {tan\,x}\) by first principle.

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Let,

\(f(x)=e\sqrt{tanx}\)

\(=\lim\limits_{h \to 0}\frac{f(x+h)-f(x)}{h}\) 

\(=\lim\limits_{h \to 0}\frac{e^\sqrt{tan(x+h)}-e^\sqrt{tanx}}{h}\) 

\(=\lim\limits_{h \to 0}e^\sqrt{tan(x+h)}\{\frac{e^\sqrt{tan(x+h)-tanx}-1}{h}\}\)\(e^\sqrt{tan(x+h)}\lim\limits_{h \to 0}\frac{e^\sqrt{tan(x+h)-tanx}-1}{\sqrt{tan(x+h)}-\sqrt{tanx}}\).\(\lim\limits_{h \to 0}\frac{e^\sqrt{tan(x+h)}-\sqrt{tanx}}{h}\)  

\(=e^\sqrt{tanx}.1.\lim\limits_{h \to 0}\frac{tan(x+h)-tanx}{h}\times\) \(\frac{1}{\sqrt{tan(x+h)}+\sqrt{tanx}}\)  

\(=e^\sqrt{tanx}.1.\lim\limits_{h \to 0}\frac{sin\,h}{h\,cos\,x(x+h)cos\,x}\times\)\(\frac{1}{\sqrt{tan(x+h)}+\sqrt{tanx}}\)

\(=\frac{e^\sqrt{tanx}}{2\sqrt{tan\,x}}.\frac{1}{cos^2x}\) 

\(=\frac{sec^2\,x}{2}.\frac{e^{tan\,x}}{\sqrt{tan\,x}}\)

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