Question:

Let $f(x) = \frac{3x^2 + 5x - 2}{2x^3 - 4x}$.

a) Find $f'(x)$ using the product rule.

b) Find $f'(x)$ using the quotient rule.

c) Evaluate $f'(2)$.

Answer:

a) To find $f'(x)$ using the product rule, we need to differentiate the numerator and denominator separately. Then, we can apply the product rule formula.

Given that $f(x) = \frac{3x^2 + 5x - 2}{2x^3 - 4x}$, we have:

$f'(x) = \frac{(3x^2 + 5x - 2)'(2x^3 - 4x) - (3x^2 + 5x - 2)(2x^3 - 4x)'}{(2x^3 - 4x)^2}$

Differentiating the numerator and denominator separately:

$(3x^2 + 5x - 2)' = 6x + 5$

$(2x^3 - 4x)' = 6x^2 - 4$

Substituting these values into the product rule formula:

$f'(x) = \frac{(6x + 5)(2x^3 - 4x) - (3x^2 + 5x - 2)(6x^2 - 4)}{(2x^3 - 4x)^2}$

Simplifying the expression:

$f'(x) = \frac{12x^4 - 24x^2 + 10x^3 - 20x - 18x^3 + 30x^2 - 12x - 24x + 8}{(2x^3 - 4x)^2}$

$f'(x) = \frac{12x^4 - 8x^3 + 6x^2 - 56x + 8}{(2x^3 - 4x)^2}$

b) To find $f'(x)$ using the quotient rule, we can directly apply the formula:

Given that $f(x) = \frac{3x^2 + 5x - 2}{2x^3 - 4x}$, we have:

$f'(x) = \frac{(3x^2 + 5x - 2)'(2x^3 - 4x) - (3x^2 + 5x - 2)(2x^3 - 4x)'}{(2x^3 - 4x)^2}$

Differentiating the numerator and denominator separately:

$(3x^2 + 5x - 2)' = 6x + 5$

$(2x^3 - 4x)' = 6x^2 - 4$

Substituting these values into the quotient rule formula:

$f'(x) = \frac{(6x + 5)(2x^3 - 4x) - (3x^2 + 5x - 2)(6x^2 - 4)}{(2x^3 - 4x)^2}$

Simplifying the expression:

$f'(x) = \frac{12x^4 - 24x^2 + 10x^3 - 20x - 18x^3 + 30x^2 - 12x - 24x + 8}{(2x^3 - 4x)^2}$

$f'(x) = \frac{12x^4 - 8x^3 + 6x^2 - 56x + 8}{(2x^3 - 4x)^2}$

c) To find $f'(2)$, we substitute $x = 2$ into the derivative $f'(x)$ obtained using either the product rule or the quotient rule.

Using the derivative obtained using the quotient rule:

$f'(2) = \frac{12(2)^4 - 8(2)^3 + 6(2)^2 - 56(2) + 8}{(2(2)^3 - 4(2))^2}$

$f'(2) = \frac{192 - 64 + 24 - 112 + 8}{(16 - 8)^2}$

$f'(2) = \frac{48}{64}$

Simplifying the expression, we get:

$f'(2) = \frac{3}{4}$

Therefore, $f'(2) = \frac{3}{4}$.