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calyne
Find dy/dx by implicit differentiation: e^(x/y) = x-y
e^(x/y) = x-y e^(xy^(-1)) = x-y differentiate, \[[-\frac{x}{y^2}\frac{dy}{dx}+\frac{1}{y}]e^{xy^{-1}}=1-\frac{dy}{dx}\] \[\frac{dy}{dx}(1-\frac{xe^{xy^{-1}}}{y^2})=1-\frac{e^{xy^{-1}}}{y}\] \[\huge \frac{dy}{dx}=\frac{1-\frac{e^{xy^{-1}}}{y}}{(1-\frac{xe^{xy^{-1}}}{y^2})}\]
multiply y^2 to top and bottom and simplify
\[e^{\frac{x}{y}}\times \frac{y-xy'}{y^2}=1-y'\] is a start
or use .sam. method either way
\[\text{Final result}\] \[\frac{dy}{dx}=\frac{y \left(e^{x/y}-y\right)}{x e^{x/y}-y^2}\]
@calyne is it clear that the right hand side becomes \[1-y'\]?
wait sam how did you get from \[[−xy2dydx+1y]exy−1=1−dydx dydx(1−xexy−1y2)=1−exy−1y\]
wait flutter i mean how did you get from the first equation to the second in your first post
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
\[[-\frac{x}{y^2}\frac{dy}{dx}+\frac{1}{y}]e^{xy^{-1}}=1-\frac{dy}{dx}\] \[-\frac{xe^{xy^{-1}}}{y^2}\frac{dy}{dx}+\frac{e^{xy^{-1}}}{y}+\frac{dy}{dx}=1\] \[\frac{dy}{dx}(1-\frac{xe^{xy^{-1}}}{y^2})=1-\frac{e^{xy^{-1}}}{y}\] \[\huge \frac{dy}{dx}=\frac{1-\frac{e^{xy^{-1}}}{y}}{(1-\frac{xe^{xy^{-1}}}{y^2})}\]
can you show me how to equate that to [ y * ( y - e^(x/y) ) ] / [ y^2 - xe^(x/y) ] ??
i got to go , bump this up for others to solve
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
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
\[e^{\frac{x}{y}}\times \frac{y-xy'}{y^2}=1-y'\] \[e^{\frac{x}{y}}(y-xy')=y^2-y^2y'\] \[ye^{\frac{x}{y}}-xe^{\frac{x}{y}}y'=y^2-y^2y'\] \[y^2y'-xe^{\frac{x}{y}}y'=y^2-ye^{\frac{x}{y}}\] \[y'(y^2-xe^{\frac{x}{y}})=y^2-ye^{\frac{x}{y}}\]
check my algebra because it is hard to write all this here, but that is the idea
last step is to divide and you are done!