RaySix

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at November 3rd 2023, 7:26:37 pm.

AP Calculus AB Exam Question:

Find the derivative of the following function using the product and quotient rules:

f(x) = \frac{(x^2 + 1)(2x - 3)}{(3x + 2)}

Step-by-step Detailed Explanation:

To find the derivative of the given function, we will use the product and quotient rules.

First, let's simplify the function by multiplying the numerator and write it as a product of two functions:

f(x) = \frac{(x^2 + 1)(2x - 3)}{(3x + 2)} = (x^2 + 1)(2x - 3) \cdot \frac{1}{(3x + 2)}

Now, applying the product rule, we can find the derivative of the first part:

\frac{d}{dx}[(x^2 + 1)(2x - 3)] = (2x - 3) \cdot \frac{d}{dx}(x^2 + 1) + (x^2 + 1) \cdot \frac{d}{dx}(2x - 3)

Simplifying, we have:

\frac{d}{dx}[(x^2 + 1)(2x - 3)] = (2x - 3)(2x) + (x^2 + 1)(2)

\frac{d}{dx}[(x^2 + 1)(2x - 3)] = 4x^2 - 6x + 2x^2 + 2

\frac{d}{dx}[(x^2 + 1)(2x - 3)] = 6x^2 - 6x + 2

Now, let's find the derivative of the second part using the quotient rule:

\frac{d}{dx}\left(\frac{1}{3x + 2}\right) = \frac{(3x + 2) \cdot \frac{d}{dx}(1) - 1 \cdot \frac{d}{dx}(3x + 2)}{(3x + 2)^2}

Simplifying, we have:

\frac{d}{dx}\left(\frac{1}{3x + 2}\right) = \frac{-1 \cdot 3}{(3x + 2)^2}

\frac{d}{dx}\left(\frac{1}{3x + 2}\right) = \frac{-3}{(3x + 2)^2}

Now, let's put the derivative of the first part and the derivative of the second part together using the quotient rule:

f'(x) = \frac{(6x^2 - 6x + 2) \cdot (3x + 2) - (x^2 + 1) \cdot (-3)}{(3x + 2)^2}

Simplifying, we have:

f'(x) = \frac{18x^3 - 6x^2 - 3x^2 - x^2 - 6x + 4x + 2 + 3}{(3x + 2)^2}

f'(x) = \frac{18x^3 - 10x^2 - 2x + 3}{(3x + 2)^2}

Therefore, the derivative of the given function is:

f'(x) = \frac{18x^3 - 10x^2 - 2x + 3}{(3x + 2)^2}