Question

Find dy/dx by implicit differentiation: e^(x/y) = x-y

Answer 1

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

Answer 2

multiply y^2 to top and bottom and simplify

Answer 3

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

Answer 4

or use .sam. method either way

Answer 5

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

Answer 6

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

Answer 7

yeah thanks guys

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

last step is to divide and you are done!