implicit differentiation problem

[racecarrr]

hi just started taking calc 1 can't figure this problem out,

differentiate: e^(x/y)=x-y

the answer given in the book is

y'=(y(y-e^(x/y)))/((y^2)-xe^(x/y))

but i just cant get there. im using the chain rule and the quotient rule but it aint workin.

any help is appreciated.

 

[BigGlenntheHeavy]

\(\displaystyle e^{x/y} \ = \ x-y\)

\(\displaystyle \frac{x}{y} \ = \ xy^{-1}. \ Now \ implicit \ differentation \ of \ xy^{-1} \ gives \ y^{-1}-xy^{-2}y'\)

\(\displaystyle Ergo \ \bigg(\frac{1}{y}-\frac{xy'}{y^2}\bigg)e^{x/y} \ = \ 1-y'\)

\(\displaystyle \frac{e^{x/y}}{y}-\frac{xy'e^{x/y}}{y^2} \ = \ 1-y'\)

\(\displaystyle y'-\frac{xy'e^{x/y}}{y^2} \ = \ 1-\frac{e^{x/y}}{y}\)

\(\displaystyle y'\bigg(1-\frac{xe^{x/y}}{y^2}\bigg) \ = \ \frac{y-e^{x/y}}{y}\)

\(\displaystyle y'\bigg(\frac{y^2-xe^{x/y}}{y^2}\bigg) \ = \ \frac{y-e^{x/y}}{y}\)

\(\displaystyle y' \ = \ \bigg(\frac{y-e^{x/y}}{y}\bigg)*\bigg(\frac{y^2}{y^2-xe^{x/y}}\bigg) \ = \ \frac{y(y-e^{x/y})}{y^2-xe^{x/y}}\)

 

[soroban]

Hello, racecarrr!

I'll do it head-on . . .

\(\displaystyle \text{Differentiate: }\;e^{\frac{x}{y}} \:=\:x-y\)

\(\displaystyle \text{Answer: }\;y'\;=\;\frac{y\left(y-e^{\frac{x}{y}}\right)} {y^2-xe^{\frac{x}{y}}}\)

\(\displaystyle \text{Differentiate implicitly: }\;e^{\frac{x}{y}}\cdit\left(\frac{y\cdot1 - x\cdot\frac{dy}{dx}}{y^2}\right) \;=\;1 - \frac{dy}{dx}\)

. . . \(\displaystyle \frac{e^{\frac{x}{y}}}{y} - \frac{xe^{\frac{x}{y}}}{y^2}\!\cdot\!\frac{dy}{dx} \;=\;1 - \frac{dy}{dx}\)

. . . \(\displaystyle \frac{dy}{dx} - \frac{xe^{\frac{x}{y}}}{y^2}\!\cdot\!\frac{dy}{dx} \;=\;1 - \frac{e^{\frac{x}{y}}}{y}\)

. . \(\displaystyle \left(1 - \frac{xe^{\frac{x}{y}}}{y^2}\right)\!\cdot\!\frac{dy}{dx} \;=\; 1 - \frac{e^{\frac{x}{y}}}{y}\)

. \(\displaystyle \left(\frac{y^2-xe^{\frac{x}{y}}}{y^2}\right)\!\cdot\!\frac {dy}{dx} \;=\;\frac{y - e^{\frac{x}{y}}}{y}\)

. . . . . . . . . . \(\displaystyle \frac{dy}{dx} \;=\;\frac{y^2}{y^2-xe^{\frac{x}{y}}}\cdot\frac{y-e^{\frac{x}{y}}}{y}\)

. . . . . . . . . . \(\displaystyle \frac{dy}{dx} \;=\;\frac{y\left(y - e^{\frac{x}{y}}\right)}{y^2-xe^{\frac{x}{y}}}\)