How do you differentiate #e^(x/y)#?

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Answer 1 · 2015-09-16T17:03:02

#d/dx(e^(x/y)) = = 3^(x/y) (y-xdy/dx)/y^2#

We use the derivative of the exponential, the chain rule and the quotient rule.

Assuming that we are to differentiate with respect to #x#, we get

#d/dx(e^(x/y)) = 3^(x/y) d/dx(x/y)#

#= 3^(x/y) (y-xdy/dx)/y^2#

We may prefer a different form, but that is all we can do without more information about how #x# and #y# are related.

If differentiating with respect to #t# the only real difference is that #d/dt(x)# may not be #1#, so we write:

#d/dt(e^(x/y)) = 3^(x/y) d/dt(x/y) = 3^(x/y) (ydx/dt-xdy/dt)/y^2#