Question

If $f\left(x\right)=x^3+7x^2+8x-9$, find f'(4).

Answer 1

Given: $f\left(x\right)=x^3+7x^2+8x-9$

Clearly, being a polynomial function, is differentiable everywhere. Therefore the derivative of $f$ at $x$ is given by:

$$\begin{gathered} f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f(x+h)-f\left(x\right)}{h} \\ \Rightarrow f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{(x+h{)}^3+7(x+h{)}^2+8(x+h)-9-x^3-7x^2-8x+9}{h} \\ \Rightarrow f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{x^3+h^3+3x^{2}h+3x{h}^2+7x^2+7h^2+14xh+8x+8h-9-x^3-7x^2-8x+9}{h} \\ \Rightarrow f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{h^3+3x^{2}h+3x{h}^2+7h^2+14xh+8h}{h} \\ \Rightarrow f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{h(h^2+3x^2+3xh+7h+14x+8)}{h} \\ \Rightarrow f\text{'}\left(x\right)=\underset{h\to 0}{\lim }{h}^2+3x^2+3xh+7h+14x+8 \\ \Rightarrow f\text{'}\left(x\right)=3x^2+14x+8 \end{gathered}$$
Thus,
$$\begin{gathered} f\text{'}\left(4\right)=3\times 4^2+14\times 4+8 \\ =48+56+8 \\ =112 \end{gathered}$$