Solutions

a) Suppose $g(x)=f(x+c)$. We have

$$\begin{align*} g'(x)&=\lim_{h\to 0}\frac{g(x+h)-g(x)}{h} \\ &=\lim_{h\to 0}\frac{f(x+c+h)-f(x+c)}{h} \\ &=f'(x+c). \end{align*}$$

b) Let $g(x)=f(cx)$. Then

$$\begin{align*} g'(x)&=\lim_{h\to 0}\frac{g(x+h)-g(x)}{h} \\ &=\lim_{h\to 0}\frac{f(c(x+h))-f(cx)}{h} \\ &=\lim_{h\to 0}\frac{f(cx+ch)-f(cx)}{h} \\ &=\lim_{h\to 0}c\left(\frac{f(cx+ch)-f(cx)}{ch}\right) \\ &=\lim_{z\to 0}c\left(\frac{f(cx+z)-f(cx)}{z}\right) \\ &=cf'(cx). \end{align*}$$

c) We have

$$\begin{align*} f'(x)&=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h} \\ &=\lim_{h\to 0}\frac{f(x+a+h)-f(x+a)}{h} \\ &=f'(x+a). \end{align*}$$