# Differentiablity proof

1. Nov 12, 2009

### John O' Meara

Suppose that a function f is differentiable at x0 and define g(x)=f(mx+b), where m and b are constants. Prove that if x1 is a point at which mx1+b=x0, then g(x) is differentiable at x1 and g'(x1)=mf'(x0).
Definition: A function f is said to be "differentiable at x0" if the limit
$$f'(x_{0})= lim_{h \rightarrow 0} \frac{f(x_{0}+h)-f(x_{0})}{h}$$
exists. If f is differentiable at each point of the open interval(a,b), then we say that f is "differentiable on (a,b)", and similarily for open intervals of the form $$(a,+\infty),(-\infty,b)and (-\infty,+\infty)$$. In the last case we say f is "differentiable everywhere". I think this is a simple proof, but since I am doing this no my own with a book that doesn't do too many example proofs, I am lost. g(x1) = f(mx1+b) = f(x0), which I think proves the limit of g(x1) exists but g'(x1)=? Thanks for the help.

2. Nov 13, 2009

### CompuChip

I think they want you to do it from the definition. I.e. show that the limit of
$$\frac{g(x_{1}+h)-g(x_{1})}{h}$$
for h to zero exists by substituting the definition of g and using that the appropriate limit for f exists.

3. Nov 13, 2009

### John O' Meara

Sorry, if I gave the wrong impression, but the definition I included myself thinking it relevant. The definition was not part of the original question.

4. Nov 14, 2009

### CompuChip

If you don't want to use the definition, you can directly prove if from the chain rule, I suppose.

I checked the proof from the definition, it works out nicely. Just remember to bring the limit you get back in the form
$$\lim_{\delta \to 0} \frac{f(x_0 + \delta) - f(x_0)}{\delta}$$
where $\delta$ is some new quantity depending on h.

(I cannot really give you more hints without doing the calculation for you, so I suggest you give this a try and post what you get).

5. Nov 14, 2009

### John O' Meara

$$\lim_{h \rightarrow 0} \frac{ g(x_{1} +h) - g(x_{1})}{h} = \lim_{h \rightarrow 0} \frac{f(m(x_{1}+h)+b) - f(mx_{1} + b)}{h}$$
$$= \lim_{h \rightarrow 0} \frac{ f(mx_{1} + mh + b) - f(mx_{1} + b)}{h}$$
$$= \lim_{mh \rightarrow 0} m\frac{ f(mx_{1}+mh+b)-f(mx_{1}+b)}{mh}$$
$$= \lim_{mh \rightarrow 0} m\frac{ f(mx_{1}+b+mh)-f(mx_{1}+b)}{mh}$$
$$= m \lim_{\delta \rightarrow 0} \frac{ f(x_{0}+ \delta)-f(x_{0})}{\delta}$$
$$= mf'(x_{0})$$

If this is ok, now I got to use this result to prove that $$f'(x)=nm(mx+b)^{n-1}$$ if $$f(x) = (mx+b)^{n}$$, where m and b are constants and n is any integer. I know that $$\frac{dx^n}{dx}=nx^{n-1}$$ where n is any integer or do I use the definition given above in #1? Thanks again.

Last edited: Nov 14, 2009
6. Nov 15, 2009

### CompuChip

Very well.

The quickest way indeed seems to be, to use that
$$\frac{d}{dx}f(x) = n x^{n-1}$$
where f(x) = xn is differentiable anywhere, and use what you have just proven.