## calyne Group Title Find dy/dx by implicit differentiation: e^(x/y) = x-y 2 years ago 2 years ago

1. .Sam. Group Title

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. Group Title

multiply y^2 to top and bottom and simplify

3. satellite73 Group Title

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

4. satellite73 Group Title

or use .sam. method either way

5. .Sam. Group Title

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

6. satellite73 Group Title

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

7. calyne Group Title

yeah thanks guys

8. calyne Group Title

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

9. calyne Group Title

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

10. .Sam. Group Title

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. Group Title

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

12. calyne Group Title

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

13. .Sam. Group Title

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

14. satellite73 Group Title

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 Group Title

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 Group Title

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

17. satellite73 Group Title

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

18. satellite73 Group Title

last step is to divide and you are done!