That is exactly the chain rule. Maybe not stated exactly as you know it.
Yeah thats what I don't understand -.-
Can you briefly explain how it is the chain rule? D:
What I was taught is dy/dx = du/dx x dy/dx, I wasn't taught the Leibniz notation.
Well you can do it using that notation but its a bit more messy IMO. (especially considering you might have to let

, if you didn't already know the derivative of an absolute function)


^See
this, if you don't understand why


}{|x^2-1|})
Now to explain why
 = g'(h(x))h'(x))
, eh? <-- probably easier said than done

In reality what you have denoted as u is what the textbook has donated as h(x).
Lets say we had,
}^2)
1.

2.
=...)
3.

4.
=...)
5.
 \times u'(x) = ...)
Alternatively...
={(2x+1)}^2)
Make this
)={(2x+1)}^2)
1.
)={h(x)}^2)
2.
}{dh(x)}=g'(h(x))=...)
3.
=2x+1)
4.
}{dx}=h'(x)=...)
5.
=g'(h(x)) \times h'(x) = ...)
Not sure you'll understand that, but hey, worth a shot
