Question

use the definition of the derivative, which is $f(x)=lim_{h \to 0}\frac{f(x + h)-f(x)}{h}$ to find the derivative of each. show all work!! 1. $f(x)=x^{2}-4x$ 2. $f(x)=x^{3}+3x^{2}-4$ 3. $f(x)=\frac{1}{x}$ 4. $f(x)=sqrt{x}$ 5. $f(x)=\frac{x + 1}{x - 2}$

Explanation:

1.

Step1: Find \(f(x + h)\)

\[

$$\begin{align*} f(x+h)&=(x + h)^2-4(x + h)\\ &=x^{2}+2xh+h^{2}-4x-4h \end{align*}$$

\]

Step2: Substitute into derivative formula

\[

$$\begin{align*} f^{\prime}(x)&=\lim_{h ightarrow0}\frac{f(x + h)-f(x)}{h}\\ &=\lim_{h ightarrow0}\frac{(x^{2}+2xh+h^{2}-4x-4h)-(x^{2}-4x)}{h}\\ &=\lim_{h ightarrow0}\frac{x^{2}+2xh+h^{2}-4x-4h - x^{2}+4x}{h}\\ &=\lim_{h ightarrow0}\frac{2xh+h^{2}-4h}{h}\\ &=\lim_{h ightarrow0}(2x + h-4) \end{align*}$$

\]

Step3: Evaluate the limit

\[

$$\begin{align*} f^{\prime}(x)&=2x-4 \end{align*}$$

\]

Step1: Find \(f(x + h)\)

\[

$$\begin{align*} f(x+h)&=(x + h)^3+3(x + h)^2-4\\ &=(x^{3}+3x^{2}h + 3xh^{2}+h^{3})+3(x^{2}+2xh+h^{2})-4\\ &=x^{3}+3x^{2}h + 3xh^{2}+h^{3}+3x^{2}+6xh+3h^{2}-4 \end{align*}$$

\]

Step2: Substitute into derivative formula

\[

$$\begin{align*} f^{\prime}(x)&=\lim_{h ightarrow0}\frac{f(x + h)-f(x)}{h}\\ &=\lim_{h ightarrow0}\frac{(x^{3}+3x^{2}h + 3xh^{2}+h^{3}+3x^{2}+6xh+3h^{2}-4)-(x^{3}+3x^{2}-4)}{h}\\ &=\lim_{h ightarrow0}\frac{x^{3}+3x^{2}h + 3xh^{2}+h^{3}+3x^{2}+6xh+3h^{2}-4 - x^{3}-3x^{2}+4}{h}\\ &=\lim_{h ightarrow0}\frac{3x^{2}h + 3xh^{2}+h^{3}+6xh+3h^{2}}{h}\\ &=\lim_{h ightarrow0}(3x^{2}+3xh + h^{2}+6x+3h) \end{align*}$$

\]

Step3: Evaluate the limit

\[

$$\begin{align*} f^{\prime}(x)&=3x^{2}+6x \end{align*}$$

\]

Step1: Find \(f(x + h)\)

\[
f(x + h)=\frac{1}{x + h}
\]

Step2: Substitute into derivative formula

\[

$$\begin{align*} f^{\prime}(x)&=\lim_{h ightarrow0}\frac{f(x + h)-f(x)}{h}\\ &=\lim_{h ightarrow0}\frac{\frac{1}{x + h}-\frac{1}{x}}{h}\\ &=\lim_{h ightarrow0}\frac{\frac{x-(x + h)}{x(x + h)}}{h}\\ &=\lim_{h ightarrow0}\frac{\frac{x - x - h}{x(x + h)}}{h}\\ &=\lim_{h ightarrow0}\frac{-h}{xh(x + h)}\\ &=\lim_{h ightarrow0}\frac{-1}{x(x + h)} \end{align*}$$

\]

Step3: Evaluate the limit

\[

$$\begin{align*} f^{\prime}(x)&=-\frac{1}{x^{2}} \end{align*}$$

\]

Answer:

\(f^{\prime}(x)=2x - 4\)

2.