9514 1404 393
Answer:
y' = -y/x
Step-by-step explanation:
It looks like you have ...
e^z -z = 0 . . . . . where z = xy
Differentiating with respect to x gives ...
z'·e^z -z' = 0
z'(e^z -1) = 0
The zero-product rule tells you this is true for e^z = 1, which is not useful for finding y', and for z' = 0, which is.
z = xy
z' = y +xy' = 0
y' = -y/x
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Additional comment
There appear to be no real values of x and y that will make your equation true.
