Q. 44 - 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.   

$f\left(x\right)={(3-x)}^2+5$

Step-by-Step Solution

Answer

The derivative of the function is $f\text{'}\left(x\right)=\begin{array}{rcl}& & 2\end{array}\begin{array}{rcl}& & x\end{array}\begin{array}{rcl}& & -\end{array}\begin{array}{rcl}& & 6\end{array}$.

Step 1. Given Information

The given function is $f\left(x\right)={(3-x)}^2+5$.

Step 2. Simplify the function

Apply the identity, ${(a-b)}^2=a^2-2ab+b^2$ in the given function.

$\begin{array}{rcl}f\left(x\right)& =& 3^2+2\left(3\right)(-x)+x^2+5\\ & =& 9-6x+x^2+5\\ & =& x^2-6x+14\end{array}$

Step 3. Find the derivative

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

$f\text{'}\left(x\right)=\frac{d}{dx}\left(x^2\right)-\frac{d}{dx}\left(6x\right)+\frac{d}{dx}\left(14\right)$

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

$\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{d}{dx}\left(x^2\right)-6\frac{d}{dx}\left(x\right)+\left(0\right)\\ & =& \frac{d}{dx}\left(x^2\right)-6\frac{d}{dx}\left(x\right)\end{array}$

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

$\begin{array}{rcl}f\text{'}\left(x\right)& =& 2x^1-6\left(1\right)\\ & =& 2x-6\end{array}$