Question

Definition-of-derivative calculations: Use the definition of the derivative to find f ' for each function

$$\begin{gathered} f\left(x\right)=x^4 \\ f\left(x\right)=x^{-2} \\ f\left(x\right)=x^2(x+1) \\ f\left(x\right)=\frac{x^2}{x+1} \end{gathered}$$

Short answer

$$\begin{gathered} f\text{'}\left(x\right)=4x^3 \\ f\text{'}\left(x\right)=-\frac{2}{x^3} \\ f\text{'}\left(x\right)=3x^2+2x \\ f\text{'}\left(x\right)=\frac{x^2+2x}{{\left(x+1\right)}^2} \end{gathered}$$

Step-by-step solution

Given Information.

Consider the given information.

$$\begin{gathered} f\left(x\right)=x^4 \\ f\left(x\right)=x^{-2} \\ f\left(x\right)=x^2(x+1) \\ f\left(x\right)=\frac{x^2}{x+1} \end{gathered}$$

Find the derivative of fx=x4.

Definition of derivative formula

$$\begin{gathered} f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f(x+h)-f\left(x\right)}{h} \\ f\left(x\right)=x^4 \\ f(x+h)={(x+h)}^4=x^4+4x^{3}h+6x^2{h}^2+4x{h}^3+h^4 \\ f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{x^4+4x^{3}h+6x^2{h}^2+4x{h}^3+h^4-x^4}{h} \\ f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{4x^{3}h+6x^2{h}^2+4x{h}^3+h^4}{h} \\ f\text{'}\left(x\right)=\underset{h\to 0}{\lim }4x^3+6x^{2}h+4x{h}^2+h^3 \\ f\text{'}\left(x\right)=4x^3 \end{gathered}$$

Find the derivative of fx=x-2.

Use the definition and find the derivative.

$$\begin{gathered} f\left(x\right)=x^{-2} \\ f(x+h)={(x+h)}^{-2}=\frac{1}{{\left(x+h\right)}^2}=\frac{1}{x^2+2xh+h^2} \\ f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{\frac{1}{x^2+2xh+h^2}-{\displaystyle \frac{1}{x^2}}}{h} \\ f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{x^2-(x^2+2xh+h^2)}{h*x^2\left(x^2+2xh+h^2\right)} \\ f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{x^2-x^2-2xh-h^2}{h*x^2\left(x^2+2xh+h^2\right)} \\ f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{-2x-h}{x^2\left(x^2+2xh+h^2\right)} \\ f\text{'}\left(x\right)=-\frac{2}{x^3} \end{gathered}$$

Find the derivative of fx=x2x+1.

Use the definition to find the derivative.

$$\begin{gathered} f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\left[\frac{f\left(x+h\right)-f\left(x\right)}{h}\right] \\ f\text{'}\left(x\right)=\underset{h\to \hspace{0.17em}0}{\lim }\left(\frac{{\left(x+h\right)}^2\left(x+h+1\right)-x^2\left(x+1\right)}{h}\right) \\ f\text{'}\left(x\right)=\underset{h\to \hspace{0.17em}0}{\lim }\left(\frac{x^3+h^2+x^2+3x^{2}h+3x{h}^2+2xh+h^3-x^3-x^2}{h}\right) \\ f\text{'}\left(x\right)=\underset{h\to \hspace{0.17em}0}{\lim }\left(\frac{h^3+h^2+3x^{2}h+3x{h}^2+2xh}{h}\right) \\ f\text{'}\left(x\right)=\underset{h\to \hspace{0.17em}0}{\lim }{h}^2+h+3x^2+3xh+2x \\ f\text{'}\left(x\right)=3x^2+2x \end{gathered}$$

Find the derivative of fx=x2x+1.

Use the definition of derivative to find it.

$$\begin{gathered} f\left(x\right)=\frac{x^2}{x+1} \\ f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\left[\frac{f\left(x+h\right)-f\left(x\right)}{h}\right] \\ =\underset{h\to \hspace{0.17em}0}{\lim }\left(\frac{\frac{{\left(x+h\right)}^2}{x+h+1}-\frac{x^2}{x+1}}{h}\right) \\ =\underset{h\to \hspace{0.17em}0}{\lim }\left(\frac{\frac{x{h}^2+x^{2}h+2xh+h^2}{\left(x+1\right)\left(x+h+1\right)}}{h}\right) \\ =\underset{h\to \hspace{0.17em}0}{\lim }\left(\frac{x{h}^2+x^{2}h+2xh+h^2}{\left(x+1\right)\left(x+h+1\right)h}\right) \\ =\underset{h\to \hspace{0.17em}0}{\lim }\left(\frac{xh+x^2+2x+h}{\left(x+1\right)\left(x+h+1\right)}\right) \\ f\text{'}\left(x\right)=\frac{x^2+2x}{{\left(x+1\right)}^2} \end{gathered}$$