Post

Created by @nathanedwards
 at November 8th 2023, 5:19:43 am.

AP Calculus AB Exam Question

Consider the function f(x) = (2x^3 + x^2 - 5x) / (x + 2).

a) Use the Product Rule to find the derivative of f(x).

b) Use the Quotient Rule to find the derivative of f(x).

Answer:

a) To find the derivative of f(x) using the Product Rule, we can set f(x) as the product of two functions: u(x) = 2x^3 + x^2 - 5x and v(x) = 1/(x + 2).

The Product Rule states that for two differentiable functions, u(x) and v(x), the derivative of their product f(x) = u(x) * v(x) is given by:

f'(x) = u'(x) * v(x) + u(x) * v'(x)

Let's find the derivatives of u(x) and v(x):

u'(x) = d/dx (2x^3 + x^2 - 5x) = 6x^2 + 2x - 5

v'(x) = d/dx (1/(x + 2)) = -1/(x + 2)^2

Now, we can use the Product Rule to find the derivative of f(x):

f'(x) = (6x^2 + 2x - 5) * (1/(x + 2)) + (2x^3 + x^2 - 5x) * (-1/(x + 2)^2)

Simplifying this expression further, we can write:

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

So, the derivative of f(x) using the Product Rule is given by (6x^2 + 2x - 5)/(x + 2) - (2x^3 + x^2 - 5x)/(x + 2)^2.

b) To find the derivative of f(x) using the Quotient Rule, we can set f(x) as the quotient of two functions: u(x) = 2x^3 + x^2 - 5x and v(x) = (x + 2).

The Quotient Rule states that for two differentiable functions, u(x) and v(x), the derivative of their quotient f(x) = u(x) / v(x) is given by:

f'(x) = (u'(x) * v(x) - u(x) * v'(x)) / v(x)^2

Let's find the derivatives of u(x) and v(x) again:

u'(x) = 6x^2 + 2x - 5

v'(x) = 1

Now, we can use the Quotient Rule to find the derivative of f(x):

f'(x) = ((6x^2 + 2x - 5) * (x + 2) - (2x^3 + x^2 - 5x) * 1) / (x + 2)^2

Expanding and simplifying this expression further, we get:

f'(x) = (6x^3 + 12x^2 + 2x^2 + 4x - 5x - 10 - 2x^3 - x^2 + 5x) / (x + 2)^2

f'(x) = (4x^3 + 14x^2 - 11) / (x + 2)^2

Therefore, the derivative of f(x) using the Quotient Rule is given by (4x^3 + 14x^2 - 11) / (x + 2)^2.

Note: It is always a good practice to simplify the final derivative expression as much as possible.