Answer
The rate of change of the function $f(x)=4x^2$ is equal to $8x$ and $\displaystyle f'(2)=16$.
Work Step by Step
\[f(x)=4x^2\]
The derivative of the function $f(x)=4x^2$ is equal to
\[f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}\]
\[\Rightarrow f'(x)=\lim_{h\rightarrow 0}\frac{4(x+h)^2-4x^2}{h}\]
\[\Rightarrow f'(x)=\lim_{h\rightarrow 0}\frac{4(x+h)^2-4x^2}{h}\]
\[\Rightarrow f'(x)=\lim_{h\rightarrow 0}\frac{4(x^2+h^2+2xh)-4x^2}{h}\]
\[\Rightarrow f'(x)=\lim_{h\rightarrow 0}\frac{4x^2+4h^2+8xh-4x^2}{h}\]
\[\Rightarrow f'(x)=\lim_{h\rightarrow 0}\frac{4h^2+8xh}{h}\]
\[\Rightarrow f'(x)=\lim_{h\rightarrow 0}\frac{h(4h+8x)}{h}\]
\[\Rightarrow f'(x)=\lim_{h\rightarrow 0}(4h+8x)=4(0)+8x=8x\]
\[f'(x)=8x\]
Substitute $x=2$
\[f'(2)=8(2)=16\]
Hence , The rate of change of the function $f(x)=4x^2$ is equal to $8x$ and $\displaystyle f'(2)=16$
