f ' (0) f(x)= [ x^2(cosx)^cotx, x is not 0 ]

Question

[ 0 , x=0 ]

Answer 1

not sure what you're after here, but if u and v are functions of x, then
d/dx u^v = v u^(v-1) v' + lnu u^v u'
Now use that and the product rule to find f'(x), if that's what you need.

Answer 2

sorry can you give me more information?

Answer 3

*sigh*
d/dx x^2(cosx)^cotx = 2x (cosx)^cotx + x^2 (v u^(v-1) v' + lnu u^v u')
where u = cosx and v = cotx

maybe you can give me some more information. All that mumbo jumbo you wrote isn't very clear.

Answer 4

f ' (0) for the function f(x)= { x^2(cosx)^cotx, x is not 0 }
{ 0 , x=0 }

And can I ask that why the website said that more foul language or rude behavior is detected......
I never said any badword before

Answer 5

When teacher oobleck said
mumbo jumbo he was trying to tell you that he doesn't understand the way you presented your question

Look what I did
And let obleck confirm more for you

F(x)=x^2(cosx)^cotx

Using product rule
U=x². V=(cosx)^cotx

U'=2x

V=(cosx)^cotx

lnv=cotx.lncos(x)

(I/v)v'=[uv'+vu'={(-sinx/cosx(cotx)-lncos(x)cosec²x}=sinx/cosx(cosx/sinx)+lncosx(cosec²x)=

V'(x)=-cosx^(cotx)lncosxcosec²x

F'(x)=uv'+vu'=-x²cosx^(cotx)(lncos(x)cosec²x+cosx^(cotx)2x
=xcos(x)^(cotx)[2-cosec²x.lncos(x)]

Answer 6

The simplified form of the last part

I omitted 'x'

But this is it

=xcos(x)^(cotx)[2-xcosec²x.lncos(x)]