Differentiation of $g(x)=f(x+c)$ 
Suppose $g(x)=f(x+c)$.
Prove $g'(x)=f'(x+c)$

I know that $$g'(x)=\lim_{x\to 0} \frac{g(x+h)-g(x)}{h}$$ that is it please help.

Fix $x$ and let $\tilde{x} = x+c$. Then
$$\lim_{h \to 0} \frac{g(x+h) - g(x)}{h} = \lim_{h \to 0} \frac{f((x+h)+c) - f(x+c)}{h} $$ $$= \lim_{h \to 0} \frac{f(\tilde{x}+h) - f(\tilde{x})}{h} = f'(\tilde{x}) = f'(x+c).$$
You could also have used the chain rule: letting $u(x) = x+ c$, we have that $g = f \circ u$, so that $g'(x) = f'(u(x)) \cdot u'(x) = f'(x+c) \cdot 1 = f'(x+c)$.
