Chain Rule - What have I done wrong?

Question

I'm having trouble with what I have done wrong with the chain rule below. I have tried to show my working as much as possible for you to better understand my issue here.

So:

Find $dy/dx$ for $y=(x^2-x)^3$
So power to the front will equal = $3(x^2-x)^2 * (2x-1)$

Where did the $-1$ come from in $2x-1$?

How did they get that?

Thanks!

Answer 1

Since $(x^n)' = n x^{n-1}$, we have $(x^2)' = 2x$ and $x' = 1$, therefore $(x^2 -x)' = 2x-1$ - differentiation of functions is additive.

Answer 2

Lets denote the following:

$y(x) = (x^2-x)^3$

$f(x)=x^2-x$

$g(x)=x^3$

than $y(x)=g(f(x))$ so by the chain rule you get that $y_x = g_x (f(x)) f_x (x)$

thus in your case $y_x = 3f(x)^2f_x(x)=2(x^2-x)^2(2x-1)$

Answer 3

So you are evidently confused in how the derivative of $x^2-x$ is computed. We know that $\frac{d}{dx}k f(x)=k\frac d{dx}f(x)$ and $\frac d{dx}(f(x)+g(x))=\frac d{dx}f(x)+\frac d{dx}g(x)$.
So $$\frac d{dx}(x^2-x)=\frac d{dx}x^2+\frac d{dx}(-x)=2x+(-1)\frac d{dx}(x)=2x+(-1)1=2x-1$$