Question Number 164366 by Zaynal last updated on 16/Jan/22
$$\frac{\boldsymbol{{d}}}{\boldsymbol{{dx}}}\:\left(\boldsymbol{{e}}^{\boldsymbol{{tan}}\left(\boldsymbol{{x}}\right)} \right) \\ $$$$\left\{\boldsymbol{{Z}}.\boldsymbol{\mathrm{A}}\right\} \\ $$
Commented by cortano1 last updated on 16/Jan/22
$$\:{y}\:=\:{e}^{\mathrm{tan}\:{x}} \\ $$$$\:\mathrm{ln}\:{y}\:=\:\mathrm{tan}\:{x} \\ $$$$\:\frac{{y}'}{{y}}\:=\:\mathrm{sec}\:^{\mathrm{2}} {x} \\ $$$$\:\frac{{dy}}{{dx}}\:=\:\mathrm{sec}\:^{\mathrm{2}} {x}\:\left({e}^{\mathrm{tan}\:{x}} \right)\: \\ $$
Answered by muneer0o0 last updated on 16/Jan/22
$$\frac{\boldsymbol{{d}}}{\boldsymbol{{dx}}}\:\left(\boldsymbol{{e}}^{\boldsymbol{{tan}}\left(\boldsymbol{{x}}\right)} \right)=\:\:\:\frac{{d}\left({e}^{\mathrm{tan}\:{x}} \right)}{{d}\left(\mathrm{tan}\:{x}\right)}×\frac{{d}\left(\mathrm{tan}\:{x}\right)}{{dx}}\:=\:{e}^{\mathrm{tan}\:{x}} \:\:\mathrm{sec}^{\mathrm{2}} {x}\: \\ $$