Q. 43 - Calculus

Question

Use the differentiation rules developed in this section to find the derivatives of the functions in Exercises 35-64. Note that it may be necessary to do some preliminary algebra before differentiating.  

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

Step-by-Step Solution

Answer

The derivative of the function is $f\text{'}\left(x\right)=\begin{array}{rcl}& & 81\end{array}\begin{array}{rcl}& & x\end{array}\begin{array}{rcl}& & 2\end{array}\begin{array}{rcl}& & +\end{array}\begin{array}{rcl}& & 108\end{array}\begin{array}{rcl}& & x\end{array}\begin{array}{rcl}& & +\end{array}\begin{array}{rcl}& & 36\end{array}$.

Step 1. Given Information

The given function is $f\left(x\right)={(3x+2)}^3$.

Step 2. Simplify the function

Use the algebraic identity, ${(a+b)}^3=a^3+b^3+3a^{2}b+3a{b}^2$ to simplify the given function.

$\begin{array}{rcl}f\left(x\right)& =& {\left(3x\right)}^3+2^3+3{\left(3x\right)}^2\left(2\right)+3\left(3x\right){\left(2\right)}^2=27x^3+8+54x^2+36x\end{array}$

Step 3. Find the derivative

• Apply the sum rule of derivative, $(f+g)\text{'}\left(x\right)=f\text{'}\left(x\right)+g\text{'}\left(x\right)$.

$\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{d}{dx}\left(27x^3\right)+\frac{d}{dx}(8)+\frac{d}{dx}(54x^2)+\frac{d}{dx}(36x)\end{array}$

• Apply the constant multiple rule, $f\text{'}\left(kx\right)=kf\text{'}\left(x\right)$ and the constant derivative function rule, $f\text{'}\left(k\right)=0$ in the derived equation.

$\begin{array}{rcl}f\text{'}\left(x\right)& =& 27\frac{d}{dx}\left(x^3\right)+\left(0\right)+54\frac{d}{dx}(x^2)+36\frac{d}{dx}(x)\\ & =& 27\frac{d}{dx}\left(x^3\right)+54\frac{d}{dx}(x^2)+36\frac{d}{dx}(x)\end{array}$

• Apply the power rule of derivative, $f\text{'}\left(x^n\right)=n{x}^{n-1}$.

$\begin{array}{rcl}f\text{'}\left(x\right)& =& 27\left(3x^2\right)+54(2x^1)+36(1)\\ & =& 81x^2+108x+36\end{array}$