<s>[INST] Find the derivative of $y = f(x^2 - 2x + 7)$ where $f'(10) = 2$ Determine the derivative if $y = f(x^2 - 2x + 7)$ and $f'(10) = 2$
Ok so honestly, I dont know how to solve this, or even know where to start. All i know is that we are given a point $(10, 2)$. But what is $f(x^2 - 2x +7)$ supposed to mean? 
 [/INST] Suppose $y=f(z)$ = well $f$ can be anything. Then we put $z=x^2-2x+7$ and get a new function $g(x)=f(x^2-2x+7)$. Some examples:
$$f(z)=z; f(x^2-2x+7)=x^2-2x+7=g(x)$$
$$f(z)=z^2+3; f(x^2-2x+7)=(x^2-2x+7)^2+3=g(x)$$
$$f(z)=\sin (z); f(x^2-2x+7)=\sin (x^2-2x+7)=g(x)$$
Now we want to see what we know about the derivative of $f(x^2-2x+7)$ when it could take any one of a myriad of forms. We use the chain rule - the derivative of $p(q(x))$ is $q'(x)p'(q(x))$ with $p=f$ and $q(x)=x^2-2x+7$ to obtain the derivative $$(2x-2)f'(x^2-2x+7)$$
Now we only know $f'(10)$, so we can only deal with $$x^2-2x+7=10$$ which reduces to $$(x-3)(x+1)=0$$
So $x=3$, or $x=-1$
With $x=3$ we get $4\times 2 =8$.
With $x=-1$ we get $-4 \times 2 = -8$.
Because the form of $f$ is undefined, we can't say more than this - even that our function is differentiable in other places.
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