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

[ 0 , x=0 ]

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.

sorry can you give me more information?

*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.

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

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)]

The simplified form of the last part

I omitted 'x'

But this is it

=xcos(x)^(cotx)[2-xcosec²x.lncos(x)]

To find f'(0), the derivative of the function f(x) at x=0, we can use the definition of the derivative or apply differentiation rules.

The function f(x) is defined as:

f(x) = [x^2 * (cos(x))^cot(x), if x is not 0]
[0 , if x = 0]

To find the derivative, we need to consider two cases separately:

Case 1: x is not 0
For x not equal to 0, we can differentiate f(x) using the product and chain rule as follows:
f'(x) = d/dx [x^2 * (cos(x))^cot(x)]
= (2x * (cos(x))^cot(x)) + (x^2 * d/dx[(cos(x))^cot(x)])
= (2x * (cos(x))^cot(x)) + (x^2 * [(cos(x))^cot(x)] * (-cot(x) * sin(x) / (cos(x) * ln(cos(x))^2)])

Case 2: x = 0
For x = 0, we can directly use the definition of the derivative at a single point:
f'(0) = lim(h->0) [(f(0+h) - f(0)) / h]
= lim(h->0) [(f(h) - f(0)) / h]
= lim(h->0) [(h^2 * (cos(h))^cot(h)) / h]
= lim(h->0) [(h * (cos(h))^cot(h))]

Notice that in this case, we can't simply substitute x=0 into the derivative expression obtained in Case 1 since it is not defined at x=0. We need to evaluate the limit as h approaches 0.

Computing the limit requires a more sophisticated method, such as L'Hopital's Rule or series expansion. Applying L'Hopital's Rule, we can differentiate the numerator and denominator separately:

f'(0) = lim(h->0) [(cos(h))^cot(h)]
= lim(h->0) [cot(h) * (cos(h))^(cot(h)) * (-sin(h)) / ln(cos(h))]
= lim(h->0) [(-sin(h) / ln(cos(h))) * (cos(h))^(cot(h) - 1)]
= (-sin(0) / ln(cos(0))) * (cos(0))^(cot(0) - 1)
= (0 / ln(1)) * (1)^(1 - 1)
= 0

Therefore, f'(0) = 0.