Q. 42 - 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. 

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

Step-by-Step Solution

Answer

The derivative of the function is $f\text{'}\left(x\right)=\begin{array}{rcl}& & 4\end{array}\begin{array}{rcl}& & x\end{array}\begin{array}{rcl}& & 3\end{array}\begin{array}{rcl}& & -\end{array}\begin{array}{rcl}& & 12\end{array}\begin{array}{rcl}& & x\end{array}$.

Step 1. Given Information

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

Step 2. Simplify the function

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

$\begin{array}{rcl}f\left(x\right)& =& {\left(x^2\right)}^2-2\left(x^2\right)\left(3\right)+3^2\\ & =& x^4-6x^2+9\end{array}$

Step 3. Find the derivative

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

$f\text{'}\left(x\right)=\frac{d}{dx}\left(x^4\right)-\frac{d}{dx}\left(6x^2\right)+\frac{d}{dx}\left(9\right)$

• 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.

$\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{d}{dx}\left(x^4\right)-6\frac{d}{dx}\left(x^2\right)+0\\ & =& \frac{d}{dx}\left(x^4\right)-6\frac{d}{dx}\left(x^2\right)\end{array}$

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

$\begin{array}{rcl}f\text{'}\left(x\right)& =& 4x^3-6\left(2x^1\right)\\ & =& 4x^3-12x\end{array}$