Suppose that the function $\displaystyle f

0, \infty) \rightarrow R$ is differentiable and let c>0. Now define $\displaystyle g

0, \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

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

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

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

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