How can we choose y and k to express f(g(x+h)) in the desired form?

[Bashyboy]

Hello everyone,

I am reading a proof of the chain rule given in this link: http://kruel.co/math/chainrule.pdf

Here is the portion I am troubled with:

"We know use these equations to rewrite f(g(x+h)). In particular, use the first equation to obtain

f(g(x+h)) = f(g(x) + [g'(x) + v]h),

and use the second equation applied to the right-hand-side with k = [g'(x) + v]h..."


How do they arrive at this, k = [g'(x) + v]h. Based above previous equations and definitions, I don't see how it is possible to write k in terms of the derivative of g(x), v, and h.

Could someone help me?

 

[pasmith]

Hello everyone,

I am reading a proof of the chain rule given in this link: http://kruel.co/math/chainrule.pdf

Here is the portion I am troubled with:

"We know use these equations to rewrite f(g(x+h)). In particular, use the first equation to obtain

f(g(x+h)) = f(g(x) + [g'(x) + v]h),

and use the second equation applied to the right-hand-side with k = [g'(x) + v]h..."How do they arrive at this, k = [g'(x) + v]h. Based above previous equations and definitions, I don't see how it is possible to write k in terms of the derivative of g(x), v, and h.

Could someone help me?

Start with:
$$f(y + k) = f(y) + k(f'(y) + w) \\ g(x + h) = g(x) + h(g'(x) + v)$$

We want to calculate $f(g(x+h)) - f(g(x))$, so first we need $f(g(x+h))$. From the second equation,
$$f(g(x + h)) = f(g(x) + h(g'(x) + v))$$
and now we apply the first equation with $y = g(x)$ and $k = h(g'(x) + v)$.

 

[HallsofIvy]

Hello everyone,

I am reading a proof of the chain rule given in this link: http://kruel.co/math/chainrule.pdf

Here is the portion I am troubled with:

"We know use these equations to rewrite f(g(x+h)). In particular, use the first equation to obtain

f(g(x+h)) = f(g(x) + [g'(x) + v]h),

and use the second equation applied to the right-hand-side with k = [g'(x) + v]h..."


How do they arrive at this, k = [g'(x) + v]h. Based above previous equations and definitions, I don't see how it is possible to write k in terms of the derivative of g(x), v, and h.

Could someone help me?

What "previous" equation or definition did they have involving k? It looks to me like they are defining k to be [g'(x)+ v]h. Had they already defined it as something else?

 

[Bashyboy]

Well, if they are defining k as [g'(x) + v]h, that would seem awfully arbitrary. What is the justification for such a definition?

 

[pasmith]

Well, if they are defining k as [g'(x) + v]h, that would seem awfully arbitrary. What is the justification for such a definition?

Go back to here:
$$f(g(x+h))=f(g(x)+h(g′(x)+v))$$
We also have
$$f(y + k) = f(y) + k(f'(y) + w))$$
which holds for all $y$ and for all $k$.

Thus, to express $f(g(x+h))$ in the form $f(g(x)) + (\mbox{something involving x and h})$, which is what we must do to attain our ultimate goal of finding $f(g(x+h))- f(g(x))$, we need to choose $y$ and $k$ subject to
$$y + k = g(x) + h(g'(x) + v).$$
What choice for $y$ and $k$ would you make here?