Chapter 2.9, Problem 22E

a.

To determine

To find derivative of the function

  $F\left(x\right)={\left(1+2x\right)}^3$

By expanding the binomial

Answer to Problem 22E

  $F^{\prime }\left(x\right)=24x^2+24x+6$

Explanation of Solution

Given:

  $F\left(x\right)={\left(1+2x\right)}^3$

Concept Used:

• Addition/subtraction rule of differentiation formula

$\frac{d}{dx}\left(f\left(x\right)\pm g\left(x\right)\right)=\frac{d}{dx}f\left(x\right)\pm \frac{d}{dx}g\left(x\right)$

• Power rule of differentiation formula

$\frac{d}{dx}{x}^n=n{x}^{n-1}$

• ${\left(a+b\right)}^3=a^3+b^3+3a^{2}b+3a{b}^2$

Calculation:

To find derivative of the function

  $F\left(x\right)={\left(1+2x\right)}^3$

First expand the binomial as

  $\begin{array}{l}F\left(x\right)={\left(1+2x\right)}^3\\ F\left(x\right)=1^3+{\left(2x\right)}^3+3{\left(2x\right)}^2\left(1\right)+3\times 1^2\times 2x\\ F\left(x\right)=1+8x^3+12x^2+6x\end{array}$

Now, differentiate it, as

  $\begin{array}{l}\frac{d}{dx}\left[F\left(x\right)\right]=\frac{d}{dx}\left(1+8x^3+12x^2+6x\right)\\ F^{\prime }\left(x\right)=\frac{d}{dx}1+\frac{d}{dx}\left(8x^3\right)+\frac{d}{dx}\left(12x^2\right)+\frac{d}{dx}\left(6x\right)\\ F^{\prime }\left(x\right)=0+8\frac{d}{dx}\left(x^3\right)+12\frac{d}{dx}{x}^2+6\frac{d}{dx}x\\ F^{\prime }\left(x\right)=8\left(3x^{3-1}\right)+12\left(2x^{2-1}\right)+6\left(1\right)\\ F^{\prime }\left(x\right)=24x^2+24x+6\end{array}$

Hence,

  $F^{\prime }\left(x\right)=24x^2+24x+6$

b.

To determine

To find derivative of the function

  $F\left(x\right)={\left(1+2x\right)}^3$

By using chain rule.

Answer to Problem 22E

  $F^{\prime }\left(x\right)=24x^2+24x+6$

Explanation of Solution

Given:

  $F\left(x\right)={\left(1+2x\right)}^3$

Concept Used:

• Chain rule of differentiation is: $\frac{d}{dx}f\left[g\left(x\right)\right]=\frac{d}{dg}f\left(g\right)\frac{dg}{dx}$ , where g is a differentiable function at x and f is a differentiable function at $g\left(x\right)$
• Power rule of differentiation formula

$\frac{d}{dx}{x}^n=n{x}^{n-1}$

Calculation:

To find derivative of the function

  $F\left(x\right)={\left(1+2x\right)}^3$

Use chain rule of differentiation that is,

  $\frac{d}{dx}f\left[g\left(x\right)\right]=\frac{d}{dg}f\left(g\right)\frac{dg}{dx}$

Where, $f\left(x\right)=x^3;g\left(x\right)=\left(1+2x\right)$

Thus, differentiation of given function can be calculated as

  $\begin{array}{l}\frac{d}{dx}\left[F\left(x\right)\right]=\frac{d}{dx}{\left(1+2x\right)}^3\\ F^{\prime }\left(x\right)=3{\left(1+2x\right)}^{3-1}\frac{d}{dx}\left(1+2x\right)\\ F^{\prime }\left(x\right)=3{\left(1+2x\right)}^2\left(\frac{d}{dx}1+2\frac{d}{dx}x\right)\\ F^{\prime }\left(x\right)=3\left(1+{\left(2x\right)}^2+4x\right)\left(0+2\left(1\right)\right)\\ F^{\prime }\left(x\right)=\left(3+12x^2+12x\right)2\\ F^{\prime }\left(x\right)=24x^2+24x+6\end{array}$

Hence,

  $F^{\prime }\left(x\right)=24x^2+24x+6$