Differentiablity proof

**John O' Meara** · Nov12-09, 09:02 PM

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
$LaTeX Code: fsingle-quote(x_{0})= lim_{h \\rightarrow 0} \\frac{f(x_{0}+h)-f(x_{0})}{h}$
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 $LaTeX Code: (a,+\\infty),(-\\infty,b)and (-\\infty,+\\infty)$ . 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.

**CompuChip** · Nov13-09, 05:24 AM

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

**John O' Meara** · Nov13-09, 12:29 PM

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.

**CompuChip** · Nov14-09, 04:45 AM

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
$LaTeX Code: \\lim_{\\delta \\to 0} \\frac{f(x_0 + \\delta) - f(x_0)}{\\delta}$
where $LaTeX Code: \\delta$ 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).

**John O' Meara** · Nov14-09, 03:46 PM

$LaTeX Code: \\lim_{h \\rightarrow 0} \\frac{ g(x_{1} +h) - g(x_{1})}{h} = \\lim_{h \\rightarrow 0} \\frac{f(m(x_{1}+h)+b) - f(mx_{1} + b)}{h}$
$LaTeX Code: = \\lim_{h \\rightarrow 0} \\frac{ f(mx_{1} + mh + b) - f(mx_{1} + b)}{h}$
$LaTeX Code: = \\lim_{mh \\rightarrow 0} m\\frac{ f(mx_{1}+mh+b)-f(mx_{1}+b)}{mh}$
$LaTeX Code: = \\lim_{mh \\rightarrow 0} m\\frac{ f(mx_{1}+b+mh)-f(mx_{1}+b)}{mh}$
$LaTeX Code: = m \\lim_{\\delta \\rightarrow 0} \\frac{ f(x_{0}+ \\delta)-f(x_{0})}{\\delta}$
$LaTeX Code: = mfsingle-quote(x_{0})$

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

**CompuChip** · Nov15-09, 04:47 AM

Very well.

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