How do you implicitly differentiate #-1=yx+(x-y-e^y)/(y+x)#?

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1 Answer
t0hierry
Mar 30, 2017

It's simple.

Explanation:

#d(-1) = 0#
#d(xy)=dx y + x dy#
#d(y+x)^(-1) = -1/(y+x)^2 (dx + dy)#
#d(x-y-e^y) = dx - dy - dy e^y#
so
#0 = y dx+ x dy -(x-y - e^y)1/(y+x)^2 (dx + dy)+1/(y+x) (dx - dy - dy e^y)#
You can continue and put all the dx and dy terms together and divide by dx. But that's going beyond what is asked, which is the differential, not the derivative.
#dy(x+(x-y-e^y)/(y+x)^2-(1+e^y)/(y+x)) = dx(-y -(x-y-e^y)/(y+x)^2 + 1/(y+x))#
This is only useful if you can separate variables.

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