Question

differentiate
f(x) = xe^2cscx

Answer 1

is this \[f(x)=xe^{2\csc x}\]?

Answer 2

i mean e^x

Answer 3

xe^xcscx

Answer 4

\[f(x)=xe^x\csc x\]?

Answer 5

yes

Answer 6

well you know the product rule, right?

Answer 7

yes

Answer 8

so split this sucker up into two parts, like\[f(x)=(xe^x)\cdot(\csc x)\]then the derivative becomes\[f'(x)=(xe^x)'(\csc x)+(xe^x)(csc x)'\]

Answer 9

\[e^xcscx+xe^x-cscxcotx\]?

Answer 10

maybe i didn't do it lol

Answer 11

hm no seems a bit off
do you follow my explanation?

Answer 12

yeah

Answer 13

i think you just made a small error somewhere. your answer is very close, just missing a csc term

Answer 14

i don't get it

Answer 15

Product rule works the same way with three terms :d
\[\Large\rm (uvw)'=u'vw+uv'w+uvw'\]
\[\large\rm (x e^x \csc x)'=(x)' e^x \csc x+x(e^x)' \csc x+ x e^x(\csc x)'\]

Answer 16

thanks

Answer 17

So your middle term just needs to be fixed a little bit :)