Answer:
[tex]h'(x)=0[/tex]
Step-by-step explanation:
By definition, the derivative of [tex]f(x)[/tex] is given by:
[tex]\displaystyle \frac{dy}{dx}=f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}[/tex]
Given [tex]h(x)=c\pi[/tex], [tex]h(x+4)[/tex] must also be [tex]c\pi[/tex]. Therefore, we have:
[tex]\displaystyle h'(x)=\lim_{h\rightarrow 0}\frac{c\pi - c\pi }{h},\\h'(x)=\lim_{h\rightarrow 0}\frac{0}{h}=\boxed{0}[/tex]
Note that the derivative of a constant is always zero since for [tex]f(x)=c[/tex], where [tex]c[/tex] is some constant, [tex]f(x)=f(x+h)[/tex] and we obtain [tex]c-c[/tex] in the numerator, yielding a final answer of zero.
