Chapter 3.3, Problem 16E

To determine

To calculate: The value of $\frac{dy}{dx}$ .

Answer to Problem 16E

The required value of $\frac{dy}{dx}$ is $7x^6+10x^4+4x^3+6x^2+2x$ .

Explanation of Solution

Given information:

The expression: $\left(x^3+x+1\right)\left(x^4+x^2+1\right)$

Formula used:

Power rule:

  $\frac{d}{dx}{x}^n=n{x}^{n-1}$

Calculation:

Consider the expression be $y=\left(x^3+x+1\right)\left(x^4+x^2+1\right)$ .

To find the required value of $\frac{dy}{dx}$ , first multiply each term of the above expression.

  $\begin{array}{c}y=\left(x^3+x+1\right)\left(x^4+x^2+1\right)\\ y=x^3\left(x^4+x^2+1\right)+x\left(x^4+x^2+1\right)+1\left(x^4+x^2+1\right)\\ y=x^7+x^5+x^3+x^5+x^3+x+x^4+x^2+1\\ y=x^7+2x^5+x^4+2x^3+x^2+1\end{array}$

Use the power rule on the power rule on the above expression.

  $\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}\left(x^7+2x^5+x^4+2x^3+x^2+1\right)\\ =\frac{d}{dx}\left(x^7\right)+\frac{d}{dx}\left(2x^5\right)+\frac{d}{dx}\left(x^4\right)+\frac{d}{dx}\left(2x^3\right)+\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(1\right)\\ =7x^{\left(7-1\right)}+10x^{\left(5-1\right)}+4x^{\left(4-1\right)}+6x^{\left(3-1\right)}+2x^{\left(2-1\right)}+0\\ =7x^6+10x^4+4x^3+6x^2+2x\end{array}$

Hence, the required value of $\frac{dy}{dx}$ is $7x^6+10x^4+4x^3+6x^2+2x$ .