Question

If $e^x(x+e)=e,$ show that $\frac{dy}{dx}=e^x+y+1$

Answer 1

$\begin{array}{c}{e}^x\left(x+e\right)=e\\ lety=e^x\left(x+1\right)-1\\ differentiatew.r.tox\\ dy/dx=\frac{d}{dx}\left[e^x\left(x+1\right)\right]-0\\ \Rightarrow dy/dx=e^x\frac{dx}{dx}\frac{xd{e}^x}{dx}+\frac{d{e}^x}{dx}\\ \Rightarrow dy/dx=e^x+x{e}^x+e^x\\ \Rightarrow \therefore dy/dx{e}^x\left[1+e^x\right]\left(x+1\right)\\ \Rightarrow dy/dx=e^x+e^x\left(x+1\right)\\ \Rightarrow dy/dx=e^x+y+1Ans.\end{array}$