Question

Use (a) the $h\to 0$definition of the derivative and then

(b) the $z\to c$definition of the derivative to find $f\text{'}\left(c\right)$ for each function f and value $x=c$ in Exercises 23–38.

28.$f\left(x\right)=x^4+1,x=2$

Short answer

$f\text{'}\left(2\right)=32$.

Step-by-step solution

Part (a) Step 1. Given information.

A function is given as$f\left(\right(x)=x^4+1$and x=c=2.

Part (a) Step 2. Solve f'(2) using h→0 definition of the derivative.

We have

$$\begin{gathered} f\text{'}\left(c\right)=\underset{h\to 0}{\lim }\frac{f(c+h)-f\left(c\right)}{h} \\ f\text{'}\left(2\right)=\underset{h\to 0}{\lim }\frac{f(2+h)-f\left(2\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{{(2+h)}^4+1-[2^4+1]}{h} \\ =\underset{h\to 0}{\lim }\frac{{(2+h)}^4+1-2^4-1}{h} \\ =\underset{h\to 0}{\lim }\frac{[{(2+h)}^2{]}^2-{\left(2^2\right)}^2}{h} \\ =\underset{h\to 0}{\lim }\frac{[{(2+h)}^2-2^2][{(2+h)}^2+2^2]}{h} \\ =\underset{h\to 0}{\lim }\frac{(h^2+4h)(h^2+4h+8)}{h} \\ =\underset{h\to 0}{\lim }(h+4)(h^2+4h+8) \\ =\underset{h\to 0}{\lim }(h^3+4h^2+4h^2+16h+8h+32) \\ =\underset{h\to 0}{\lim }(h^3+8h^2+24h+32) \\ =0+0+0+32 \\ =32 \end{gathered}$$

Part (b) Step 1. Solve f'(2) using z→2 definition of the derivative.

We have

$$\begin{gathered} f\text{'}\left(c\right)=\underset{z\to c}{\lim }\frac{f\left(z\right)-f\left(c\right)}{z-c} \\ f\text{'}\left(2\right)=\underset{z\to 2}{\lim }\frac{f\left(z\right)-f\left(2\right)}{z-2} \\ =\underset{z\to 2}{\lim }\frac{z^4+1-(2^4+1)}{z-2} \\ =\underset{z\to 2}{\lim }\frac{z^4-2^4}{z-2} \\ =\underset{z\to 2}{\lim }\frac{(z^2-2^2)(z^2+2^2)}{z-2} \\ =\underset{z\to 2}{\lim }\frac{(z-2)(z+2)(z^2+4)}{z-2} \\ =\underset{z\to 2}{\lim }(z+2)(z^2+4) \\ =\underset{z\to 2}{\lim }(z^3+4z+2z^2+8) \\ =2^3+4\left(2\right)+2{\left(2\right)}^2+8 \\ =8+8+8+8 \\ =32 \end{gathered}$$