Chapter 3.9, Problem 1E

To determine

To find the value of $f\text{'}(x)$ if the function is $f(x)=\frac{4}{3x^2+2}.$

Answer to Problem 1E

  $\frac{-24x}{{\left(3x^2+2\right)}^2}$

Explanation of Solution

Given:

Given function: $f(x)=\frac{4}{3x^2+2}$

Concept used:

Derivative of function is defined as

  $f\text{'}(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$

Calculation:

  $\begin{array}{l}\Rightarrow f(x)=\frac{4}{3x^2+2}\\ \Rightarrow f(x+h)=\frac{4}{3{(x+h)}^2+2}\end{array}$

Calculating derivative,

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

  $\Rightarrow f\text{'}(x)=\underset{h\to 0}{\lim }\frac{\frac{12x^2+8-4(3(x^2+h^2+2xh)+2)}{\left(3{(x+h)}^2+2\right)\left(3x^2+2\right)}}{h}$

  $\Rightarrow f\text{'}(x)=\underset{h\to 0}{\lim }\frac{\frac{12x^2+8-4(3x^2+3h^2+6xh+2)}{\left(3{(x+h)}^2+2\right)\left(3x^2+2\right)}}{h}$

  $\Rightarrow f\text{'}(x)=\underset{h\to 0}{\lim }\frac{\frac{\cancel{12x^2}+\cancel{8}-\cancel{12x^2}-12h^2-24xh-\cancel{8}}{\left(3{(x+h)}^2+2\right)\left(3x^2+2\right)}}{h}$

  $\Rightarrow f\text{'}(x)=\underset{h\to 0}{\lim }\frac{\frac{-12h^2-24xh}{\left(3{(x+h)}^2+2\right)\left(3x^2+2\right)}}{h}$

  $\Rightarrow f\text{'}(x)=\underset{h\to 0}{\lim }\frac{\frac{-\cancel{h}(12h+24x)}{\left(3{(x+h)}^2+2\right)\left(3x^2+2\right)}}{\cancel{h}}$

  $\Rightarrow f\text{'}(x)=\underset{h\to 0}{\lim }\frac{-(12h+24x)}{\left(3{(x+h)}^2+2\right)\left(3x^2+2\right)}$

  $\Rightarrow f\text{'}(x)=\frac{-(12(0)+24x)}{\left(3{(x+0)}^2+2\right)\left(3x^2+2\right)}$

  $\Rightarrow f\text{'}(x)=\frac{-(24x)}{\left(3{(x)}^2+2\right)\left(3x^2+2\right)}$

  $\Rightarrow f\text{'}(x)=\frac{-24x}{\left(3x^2+2\right)\left(3x^2+2\right)}$

  $\Rightarrow f\text{'}(x)=\frac{-24x}{{\left(3x^2+2\right)}^2}$

Conclusion:

Therefore, derivative of $f(x)=\frac{4}{3x^2+2}$ is $\frac{-24x}{{\left(3x^2+2\right)}^2}.$