Derivative of sin(x)^2*cos(x)

Your notation is wrong, hence is your confusion, perhaps it should look as (sinx)^2*(cosx):((sinx)^2*(cosx))' = ((sin x)^2)'*cos x + (sin x)^2*(cos x)' = [2sin x cos x] *cos x - (sin x)^2*sin x =2sin x cos ^2 x - (sin x)^3

Really missed something, did that stuff long ago.

Thanks for your reply! Though, I checked the notation. Maybe it's more clear with superscript: f(x) = sin ² (x)* cos(x) and it's derivative is supposed to be: f'(x) = 2*sin(x) * cos ²(x) -sin ³(x)

It's actually part of a limit where you use the rule of l'Hopital. I don't know if that helps. Any suggestion is welcome!

Frankly, it mudded waters completely. L'Hopital rule is used when indeterminacy occurs in order to evaluate it. I guess that expression is a part of the task. Could you present the whole of it?

[input]diff( sin(x)^2*cos(x), x)[/input]

f(x) = sin ² (x)* cos(x)

Are you familiar with the product rule?
the product rule says that if
y = uv
then y' = uv' + vu'

let
u = sin ² (x) = (sin x) ²
u' = 2* sin x * cos x (I have used composite function rule or chain rule to get this derivative)

v = cos x
v' = -sin x

so
y' = [ (sin x) ²]*[-sin x] + [cos x ] * [2* sin x * cos x]

y' = -(sin x) ³ + 2 *sin x * (cos x) ²

y' = 2 *sin x * (cos x) ² -(sin x) ³

y' = 2*sin x*cos ²x - sin ³x

Do you understand what I have done?