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Question Number 29 by user1 last updated on 25/Jan/15
Differentiate   e^(√(cot x)) .
$$\mathrm{Differentiate}\:\:\:{e}^{\sqrt{\mathrm{cot}\:{x}}} . \\ $$
Answered by user2 last updated on 03/Nov/14
Let   y=e^(√(cot x))   Put cot x =t and (√(cot x))=(√t)=u, so that                      y=e^u   (dy/du)=e^u ,   (du/dt)= (1/2)t^(−1/2)   = (1/(2(√t)))   and  (dt/dx)= −cosec^2 x  so,  ((dy/dx)=(dy/du)×(du/dt)×(dt/dx)) =−(1/2)×((cosec^2 x)/(2(√t)))e^u   =((−cosec^2 x)/(2(√t)))×e(√t)               [∵ u=(√t) ]  =((−cosec^2 x)/(2(√(cot x))))×e^(√(cot x))
$$\mathrm{Let}\:\:\:{y}={e}^{\sqrt{\mathrm{cot}\:{x}}} \\ $$$$\mathrm{Put}\:\mathrm{cot}\:{x}\:={t}\:\mathrm{and}\:\sqrt{\mathrm{cot}\:{x}}=\sqrt{{t}}={u},\:\mathrm{so}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{y}={e}^{{u}} \\ $$$$\frac{{dy}}{{du}}={e}^{{u}} ,\:\:\:\frac{{du}}{{dt}}=\:\frac{\mathrm{1}}{\mathrm{2}}{t}^{−\mathrm{1}/\mathrm{2}} \:\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{{t}}}\: \\ $$$$\mathrm{and}\:\:\frac{{dt}}{{dx}}=\:−\mathrm{cosec}^{\mathrm{2}} {x} \\ $$$$\mathrm{so},\:\:\left(\frac{{dy}}{{dx}}=\frac{{dy}}{{du}}×\frac{{du}}{{dt}}×\frac{{dt}}{{dx}}\right)\:=−\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{cosec}^{\mathrm{2}} {x}}{\mathrm{2}\sqrt{{t}}}{e}^{{u}} \\ $$$$=\frac{−\mathrm{cosec}^{\mathrm{2}} {x}}{\mathrm{2}\sqrt{{t}}}×{e}\sqrt{{t}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\because\:{u}=\sqrt{{t}}\:\right] \\ $$$$=\frac{−\mathrm{cosec}^{\mathrm{2}} {x}}{\mathrm{2}\sqrt{\mathrm{cot}\:{x}}}×{e}^{\sqrt{\mathrm{cot}\:{x}}} \\ $$