Question

if f(x)=sin(e^-x), then f'(x) = ??

Answer 1

chain rule to the rescue
\[(f(g))'=f'(g)\times g'\] with
\[f(x)=\sin(x), f'(x)=\cos(x), g(x)=e^{-x}, g'(x)=-e^{-x}\]

Answer 2

hmm

Answer 3

i thought e^-x stays e^-x in g'(X)

Answer 4

since e^x always stays the same

Answer 5

no you need the chain rule for that one as well
\[\frac{d}{dx}[e^x]=e^{x}\] but
\[\frac{d}{dx}[e^{-x}=e^{-x}\times \frac{d}{dx}[-x]=-e^{-x}\]

Answer 6

it is easy to forget the chain rule for a constant multiple, but you still need it
for example
\[\frac{d}{dx}[\sin(2x)=2\cos(2x)\] and
\[\frac{d}{dx}[e^{-5x}]=-5e^{-5x}\]

Answer 7

so the answer is -e^-x(cos(e^-x)) ?

Answer 8

yes

Answer 9

thanks :)

Answer 10

yw

Answer 11

what is it comes the derivative of xe^lnx^2 ? double power with e