Question

compute the derivative of the given function in two different ways.
h(x)=\frac{x^{3}-8x^{2}}{3x^{2}}
a) use the quotient rule, (left\frac{f(x)}{g(x)}
ight=\frac{g(x)cdot f(x)-f(x)cdot g(x)}{g(x)^{2}}). (fill in each blank, then simplify.)
h(x)=(square)cdot(square)-(square)cdot(square)/(square)=square
b) use algebra first to simplify (h), then differentiate without the quotient rule.
h(x)=square

Explanation:

Step1: Identify f(x) and g(x) for Quotient Rule

Let $f(x)=x^{3}-8x^{2}$ and $g(x) = 3x^{2}$. Then $f'(x)=3x^{2}-16x$ and $g'(x)=6x$.

Step2: Apply Quotient Rule

$h'(x)=\frac{g(x)\cdot f'(x)-f(x)\cdot g'(x)}{[g(x)]^{2}}=\frac{(3x^{2})\cdot(3x^{2}-16x)-(x^{3}-8x^{2})\cdot(6x)}{(3x^{2})^{2}}$
$=\frac{9x^{4}-48x^{3}-6x^{4}+48x^{3}}{9x^{4}}=\frac{3x^{4}}{9x^{4}}=\frac{1}{3}$

Step3: Simplify h(x) first for part b

$h(x)=\frac{x^{3}-8x^{2}}{3x^{2}}=\frac{x^{3}}{3x^{2}}-\frac{8x^{2}}{3x^{2}}=\frac{1}{3}x-\frac{8}{3}$

Step4: Differentiate the simplified h(x)

Since the derivative of $\frac{1}{3}x$ is $\frac{1}{3}$ and the derivative of a constant $-\frac{8}{3}$ is 0, $h'(x)=\frac{1}{3}$

Answer:

a) $h'(x)=\frac{(3x^{2})\cdot(3x^{2}-16x)-(x^{3}-8x^{2})\cdot(6x)}{(3x^{2})^{2}}=\frac{1}{3}$
b) $h'(x)=\frac{1}{3}$