Question Number 29 by user1 last updated on 25/Jan/15
$$\mathrm{Differentiate}\:\:\:{e}^{\sqrt{\mathrm{cot}\:{x}}} . \\ $$
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))](Q30.png)
$$\mathrm{Let}\:\:\:{y}={e}^{\sqrt{\mathrm{cot}\:{x}}} \\ $$$$\mathrm{Put}\:\mathrm{cot}\:{x}\:={t}\:\mathrm{and}\:\sqrt{\mathrm{cot}\:{x}}=\sqrt{{t}}={u},\:\mathrm{so}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{y}={e}^{{u}} \\ $$$$\frac{{dy}}{{du}}={e}^{{u}} ,\:\:\:\frac{{du}}{{dt}}=\:\frac{\mathrm{1}}{\mathrm{2}}{t}^{−\mathrm{1}/\mathrm{2}} \:\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{{t}}}\: \\ $$$$\mathrm{and}\:\:\frac{{dt}}{{dx}}=\:−\mathrm{cosec}^{\mathrm{2}} {x} \\ $$$$\mathrm{so},\:\:\left(\frac{{dy}}{{dx}}=\frac{{dy}}{{du}}×\frac{{du}}{{dt}}×\frac{{dt}}{{dx}}\right)\:=−\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{cosec}^{\mathrm{2}} {x}}{\mathrm{2}\sqrt{{t}}}{e}^{{u}} \\ $$$$=\frac{−\mathrm{cosec}^{\mathrm{2}} {x}}{\mathrm{2}\sqrt{{t}}}×{e}\sqrt{{t}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\because\:{u}=\sqrt{{t}}\:\right] \\ $$$$=\frac{−\mathrm{cosec}^{\mathrm{2}} {x}}{\mathrm{2}\sqrt{\mathrm{cot}\:{x}}}×{e}^{\sqrt{\mathrm{cot}\:{x}}} \\ $$