Question

what is the derivative of 1+ln xy = e^(x-y)

Answer 1

Are you familiar with the chain rule? Because you want to use that here.

Answer 2

yes

Answer 3

Chain rule states:
Suppose that you have two functions f(x) and g(x) and they are both differentiable, thus:
\[F(x) = (f og)(x)\] and therefore,: \[F'(x)= f'(g(x)) \times g'(x)\] OR similarly: \[\frac{ dy }{ dx } = \frac{ dy }{ du } \frac{ du }{ dx }\]

Answer 4

It might also help to know the derivative of ln(x): \[\frac{ d }{ dx }\ln(x)=\frac{ 1 }{ x }\] and \[\frac{ d }{ dx }e^x= e^x\]

Answer 5

In this problem, you are also using product rule.
[HINT: xy]

Answer 6

my main concern is the deriv of e^(x-y). would it be e^(x-y) * y'?

Answer 7

Yes, that's correct. I'm assuming you are trying to solve for \[\frac{ dy }{ dx }\]
So you want to use algebra to rearrange it to get: \[\frac{ dy }{ dx } = f(g(x))\]

Answer 8

yeah i am