Question

Find

by using implicit differentiation:
e^x/y = 3x - y

Can someone help me solve this problem !

Answer 1

now, the assumption is that, "x" is a simple variable whilst "y" is a function in x-terms, so

`e^{(x)/(y)}=3x-y\implies \stackrel{\textit{\LARGE chain~rule}}{e^{(x)/(y)}\stackrel{quotient~rule}{\left( \cfrac{1\cdot y-x\cdot (dy)/(dx)}{y^2} \right)}} =3-\cfrac{dy}{dx} \\\\\\ \cfrac{ e^{(x)/(y)}y-e^{(x)/(y)}x\cdot (dy)/(dx)}{y^2}=3-\cfrac{dy}{dx}\implies e^{(x)/(y)}y-e^{(x)/(y)}x\cdot \cfrac{dy}{dx}=3y^2-y^2\cfrac{dy}{dx}`

`e^{(x)/(y)}y-3y^2=e^{(x)/(y)}x\cdot \cfrac{dy}{dx}-y^2\cfrac{dy}{dx}\implies e^{(x)/(y)}y-3y^2= \cfrac{dy}{dx}\left( e^{(x)/(y)}x-y^2 \right) \\\\[-0.35em] ~\dotfill\\\\ ~\hfill {\LARGE \begin{array}{llll} \cfrac{e^{(x)/(y)}y-3y^2}{e^{(x)/(y)}x-y^2}=\cfrac{dy}{dx} \end{array}}~\hfill`