Question

Let f(x)=-4e^(xcosx) f'(x)=? I need help with this problem, not sure if i am supposed to use chain rule or product rule, if so what are some ways I can recognize which rules to use on such problems?

Answer 1

the chain rule will come first, then you'll use the product rule to derive x*cos(x)

Answer 2

it might help to replace x*cos(x) with something like w, so w = x*cos(x)

f(x) = -4e^(x*cos(x))

f(x) = -4e^(w)

f ' (x) = -4e^(w) * dw/dx

when you get to dw/dx, you'll use the product rule

Answer 3

The general rule is to start on the outside and work your way inside (in terms of functions)

Answer 4

thank you @jim_thompson5910 So when using the chain rule, I take the derivative of outside and multiply it with the derivative of the inside?

Answer 5

correct

Answer 6

and you can have it nested as deep as you want (you don't necessarily need to have 2 levels deep)

Answer 7

example:

e^(sin(2x))

outermost function: f(x) = e^x
derivative: f ' (x) = e^x

middle function: g(x) = sin(x)
derivative: g ' (x) = cos(x)

innermost function: h(x) = 2x
derivative: h ' (x) = 2

-------------------------

so...

p(x) = f(g(h(x))) = e^(sin(2x))

p ' (x) = f' (g(h(x))) * g ' (h(x)) * h ' (x)

p ' (x) = e^(sin(2x)) * cos(2x) * 2

p ' (x) = 2*e^(sin(2x)) * cos(2x)

this is a chain rule example 3 levels deep (so to speak). Again, you can go as deep as you want/need to.

Answer 8

So would the derivative of the outside be -sinx ?

Answer 9

Got it, thanks for the explanation !! @jim_thompson5910