Proving a derivative exists If f is differentiable at $x$ then for $\alpha\neq1$ $$f'(x)=\lim_{c\to 0} {{f(x+c)-f(x+\alpha c)}\over{c-\alpha c}}.$$  I am not really sure what it is I need to even show to prove the statement.

Using just the derivative definition:
$$ \lim_{c\to0}{{f(x+c)-f(x+\alpha c)}\over{c-\alpha c}}=\frac{1}{1-\alpha}\lim_{c\to0}{{f(x+c)-f(x)}\over{c}}-\frac{\alpha}{1-\alpha}\lim_{\alpha c\to0}{{f(x+\alpha c)-f(x)}\over{\alpha c}}\\=\frac{f'(x)}{1-\alpha}-\frac{\alpha f'(x)}{1-\alpha}=f'(x)
$$
