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InterviewSolution
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Difference \(e\sqrt {tan\,x}\) by first principle. |
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Answer» 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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