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Calc 1: dy/dx (e^y)(cosx) I got (e^y)(-sinx) + (cosx)(e^y), but I feel I'm missing something.
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This is implicit differentiation. If this were e^x(cosx) then this would be e^x(cosx) -e^x(sinx) which is of the same form of your solution. Hint: When you take the derivative of e^y it is (e^y)y'
Yes, u = e^y -> u' = y' e^y v = cosx -> v' = ..
Thank you, That's what I thought, but could you explain why because I'm still a little confused.
Y is function of x, so when you take dy/dx ( with respect to x): chain rule need to be applied for y!
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