# The Chain Rule in n Dimensions ... Browder Theorem 8.15 - Another Question ...

#### Peter

##### Well-known member
MHB Site Helper
I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ...

I am currently reading Chapter 8: Differentiable Maps and am specifically focused on Section 8.2 Differentials ... ...

I need some further help in order to fully understand the proof of Theorem 8.15 ...

Theorem 8.15 and its proof read as follows:  In the above proof by Browder we read the following:

" ... ... Now

$$\displaystyle (g \circ f) (p + h) - (g \circ h) (p) = g(q + k) - g(q)$$ ... ... "

Can someone please show how/why $$\displaystyle (g \circ f) (p + h) - (g \circ h) (p) = g(q + k) - g(q)$$ ... ...

Help will be appreciated ...

Peter

#### Peter

##### Well-known member
MHB Site Helper
I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ...

I am currently reading Chapter 8: Differentiable Maps and am specifically focused on Section 8.2 Differentials ... ...

I need some further help in order to fully understand the proof of Theorem 8.15 ...

Theorem 8.15 and its proof read as follows:

In the above proof by Browder we read the following:

" ... ... Now

$$\displaystyle (g \circ f) (p + h) - (g \circ h) (p) = g(q + k) - g(q)$$ ... ... "

Can someone please show how/why $$\displaystyle (g \circ f) (p + h) - (g \circ h) (p) = g(q + k) - g(q)$$ ... ...

Help will be appreciated ...

Peter

It now appears to me that the answer to my question is quite straightforward ...

We have ...

$$\displaystyle (g \circ f) (p + h) - (g \circ h) (p) = g(q + k) - g(q)$$

$$\displaystyle = g[ f(p + h) ] - g[f(p)]$$ ...

But ... we have that ...

$$\displaystyle k = f(p+h) - f(p)$$

so that

$$\displaystyle f(p+h) = k + f(p) = q + k$$

Therefore

$$\displaystyle g[ f(p + h) ] - g[f(p)] = g(q + k ) - g(q)$$ ...

Don't know why I didn't see it ...

Peter