## calyne 3 years ago Find dy/dx by implicit differentiation: e^(x/y) = x-y

1. .Sam.

e^(x/y) = x-y e^(xy^(-1)) = x-y differentiate, $[-\frac{x}{y^2}\frac{dy}{dx}+\frac{1}{y}]e^{xy^{-1}}=1-\frac{dy}{dx}$ $\frac{dy}{dx}(1-\frac{xe^{xy^{-1}}}{y^2})=1-\frac{e^{xy^{-1}}}{y}$ $\huge \frac{dy}{dx}=\frac{1-\frac{e^{xy^{-1}}}{y}}{(1-\frac{xe^{xy^{-1}}}{y^2})}$

2. .Sam.

multiply y^2 to top and bottom and simplify

3. satellite73

$e^{\frac{x}{y}}\times \frac{y-xy'}{y^2}=1-y'$ is a start

4. satellite73

or use .sam. method either way

5. .Sam.

$\text{Final result}$ $\frac{dy}{dx}=\frac{y \left(e^{x/y}-y\right)}{x e^{x/y}-y^2}$

6. satellite73

@calyne is it clear that the right hand side becomes $1-y'$?

7. calyne

yeah thanks guys

8. calyne

wait sam how did you get from $[−xy2dydx+1y]exy−1=1−dydx dydx(1−xexy−1y2)=1−exy−1y$

9. calyne

wait flutter i mean how did you get from the first equation to the second in your first post

10. .Sam.

There's 2 steps there, 1)multiply e^{xy^{-1}} into [−xy2dydx+1y] 2)moving the dy/dx from RHS to LHS then factor dy/dx

11. .Sam.

$[-\frac{x}{y^2}\frac{dy}{dx}+\frac{1}{y}]e^{xy^{-1}}=1-\frac{dy}{dx}$ $-\frac{xe^{xy^{-1}}}{y^2}\frac{dy}{dx}+\frac{e^{xy^{-1}}}{y}+\frac{dy}{dx}=1$ $\frac{dy}{dx}(1-\frac{xe^{xy^{-1}}}{y^2})=1-\frac{e^{xy^{-1}}}{y}$ $\huge \frac{dy}{dx}=\frac{1-\frac{e^{xy^{-1}}}{y}}{(1-\frac{xe^{xy^{-1}}}{y^2})}$

12. calyne

can you show me how to equate that to [ y * ( y - e^(x/y) ) ] / [ y^2 - xe^(x/y) ] ??

13. .Sam.

i got to go , bump this up for others to solve

14. satellite73

if we can start with $e^{\frac{x}{y}}\times \frac{y-xy'}{y^2}=1-y'$ we can do it step by step, it is essentially algebra from here on in

15. satellite73

i would a) multiply both sides by $$y^2$$ b) put everything with a $$y'$$ on one side and everything else on the other c) factor $$y'$$ out of the terms d) divide

16. satellite73

$e^{\frac{x}{y}}\times \frac{y-xy'}{y^2}=1-y'$ $e^{\frac{x}{y}}(y-xy')=y^2-y^2y'$ $ye^{\frac{x}{y}}-xe^{\frac{x}{y}}y'=y^2-y^2y'$ $y^2y'-xe^{\frac{x}{y}}y'=y^2-ye^{\frac{x}{y}}$ $y'(y^2-xe^{\frac{x}{y}})=y^2-ye^{\frac{x}{y}}$

17. satellite73

check my algebra because it is hard to write all this here, but that is the idea

18. satellite73

last step is to divide and you are done!