1. Is this correct

Suppose that the function $\displaystyle f0, \infty) \rightarrow R$ is differentiable and let c>0. Now define $\displaystyle g0, \infty) \rightarrow R$ by $\displaystyle g(x) = f(cx)$ for x>0. Just using the definition of derivative, show that $\displaystyle g'(x) = c f'(x)$ for x>0

My work:

according to the defnition of derivative, g(x) is diffrentiable if the following limit exists

$\displaystyle \lim_{x \to {x_0}} \frac{g(x)-g(x_0)}{x-{x_0}}$

$\displaystyle \lim_{x \to {x_0}} \frac{f(cx)-f(cx_0)}{x-{x_0}}$

$\displaystyle c \lim_{x \to {x_0}} \frac{f(cx)-f(cx_0)}{c(x-{x_0})}$

$\displaystyle c \times f'(cx)$

2. Originally Posted by serious331
Suppose that the function $\displaystyle f0, \infty) \rightarrow R$ is differentiable and let c>0. Now define $\displaystyle g0, \infty) \rightarrow R$ by $\displaystyle g(x) = f(cx)$ for x>0. Just using the definition of derivative, show that $\displaystyle g'(x) = c f'(x)$ for x>0

My work:

according to the defnition of derivative, g(x) is diffrentiable if the following limit exists

$\displaystyle \lim_{x \to {x_0}} \frac{g(x)-g(x_0)}{x-{x_0}}$

$\displaystyle \lim_{x \to {x_0}} \frac{f(cx)-f(cx_0)}{x-{x_0}}$

$\displaystyle c \lim_{x \to {x_0}} \frac{f(cx)-f(cx_0)}{c(x-{x_0})}$

$\displaystyle c \times f'(cx)$

Very nice. One step before the last one though, I'd rather write:

$\displaystyle \lim_{cx\to cx_0}\,c\,\frac{f(cx)-f(cx_0)}{cx-cx_0}$

Tonio

3. Originally Posted by tonio
Very nice. One step before the last one though, I'd rather write:

$\displaystyle \lim_{cx\to cx_0}\,c\,\frac{f(cx)-f(cx_0)}{cx-cx_0}$

Tonio
Hi tonio,

I have actually written that in my notebook, but was too lazy to write that line in this thread. Thanks for your time!