Problem 16 - Calculus - AP Edition

Question

$$\text { Find } f^{\prime}(x)$$. $$f(x)=(2 \sqrt{x}+1)\left(\frac{2-x}{x^{2}+3 x}\right)$$

Step-by-Step Solution

Answer

Use the product rule and quotient rule to differentiate, resulting in a complex rational expression for \( f'(x) \).

Step 1: Identify the Function Types

The function given is a product of two functions: \( u(x) = 2\sqrt{x} + 1 \) and \( v(x) = \frac{2-x}{x^2 + 3x} \). We need to use the product rule to find \( f'(x) \).

Step 2: Recall the Product Rule

The product rule for derivatives states that if \( f(x) = u(x) \cdot v(x) \), then \( f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) \). We'll apply this rule to \( f(x) \).

Step 3: Differentiate \( u(x) \)

\( u(x) = 2\sqrt{x} + 1 \) can be rewritten as \( u(x) = 2x^{1/2} + 1 \). Using the power rule, \( u'(x) = 2 \cdot \frac{1}{2}x^{-1/2} = x^{-1/2} = \frac{1}{\sqrt{x}} \).

Step 4: Differentiate \( v(x) \) (Quotient Rule)

Apply the quotient rule to \( v(x) = \frac{2-x}{x^2 + 3x} \). If \( v(x) = \frac{g(x)}{h(x)} \), then \( v'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} \). Here, \( g(x) = 2-x \) and \( h(x) = x^2 + 3x \), so we first find their derivatives: \( g'(x) = -1 \) and \( h'(x) = 2x + 3 \). Substitute these into the quotient rule formula.

Step 5: Apply the Quotient Rule

Substitute the derivatives into the quotient rule: \[ v'(x) = \frac{-1(x^2 + 3x) - (2-x)(2x + 3)}{(x^2 + 3x)^2} \]. Simplify the numerator: \[ -x^2 - 3x - (4x - 2x^2 + 6 - 3x) = x^2 - 7x - 6 \].Thus, \[ v'(x) = \frac{x^2 - 7x - 6}{(x^2 + 3x)^2} \].

Step 6: Substitute into the Product Rule

Now substitute \( u'(x) \), \( v(x) \), \( u(x) \), and \( v'(x) \) into the product rule:\[ f'(x) = \left( \frac{1}{\sqrt{x}} \right) \left( \frac{2-x}{x^2 + 3x} \right) + (2\sqrt{x} + 1) \left( \frac{x^2 - 7x - 6}{(x^2 + 3x)^2} \right) \].

Step 7: Simplify the Expression

Simplify each term separately and then combine. You'll have: \[ f'(x) = \frac{2-x}{x^{3/2}(x^2 + 3x)} + \frac{(2\sqrt{x} + 1)(x^2 - 7x - 6)}{(x^2 + 3x)^2} \].

Key Concepts

Understanding the Chain Rule

The chain rule is a fundamental tool in calculus used to differentiate composite functions. It essentially allows us to differentiate a function that is composed of one or more functions. This is particularly useful when dealing with functions within functions, or whenever the argument of the function also involves some variable.
When applying the chain rule, imagine you have a composite function like \( f(g(x)) \). To find the derivative, you first take the derivative of the outer function \( f \), then multiply it by the derivative of the inner function \( g \). Mathematically, it is presented as:

• \( (f(g(x)))' = f'(g(x)) \cdot g'(x) \)

Remember, it's the linkage of two derivatives - hence the 'chain'. This concept is crucial for handling more complex derivatives that involve nested functions. It comes in handy in various real-world applications, such as when you deal with acceleration in physics as a derivative of velocity, when both position and velocity are functions of time.
Though not explicitly used in every step of our example, understanding the chain rule deepens your comprehension of how derivatives interlink across different levels of composed functions.

Applying the Quotient Rule

The quotient rule helps you differentiate functions that are divided by each other. In the calculus journey, it's common to encounter situations where one function is over another, forming a quotient. When you're asked to differentiate something like \( \frac{g(x)}{h(x)} \), the quotient rule is your best friend.
To apply the quotient rule, remember this formula:

• \( \left(\frac{g(x)}{h(x)}\right)' = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} \)

The derivation captures the essence of how changes in both the numerator \( g(x) \) and the denominator \( h(x) \) work together. You first find the derivative of the top function multiplied by the bottom function. Then, subtract the product of the top function and the derivative of the bottom one. Finally, the whole expression is divided by the square of the bottom function.
In the example, we applied the quotient rule to differentiate \( v(x) = \frac{2-x}{x^2+3x} \), where \( g(x) = 2-x \) and \( h(x) = x^2+3x \). After determining \( g'(x) \) and \( h'(x) \), we plugged these into the quotient rule formula to find \( v'(x) \). This meticulous process is vital to ensure the readiness for more advanced calculus challenges.

Utilizing the Power Rule in Differentiation

The power rule is one of the simplest and most often used rules in differentiation. Whenever you're dealing with functions of the form \( x^n \), this rule makes finding the derivative straightforward and less time-consuming.
To apply the power rule, if \( f(x) = x^n \), then its derivative \( f'(x) \) is computed as:

• \( f'(x) = nx^{n-1} \)

This shows that you bring down the power as a coefficient and then subtract one from the original power. This handy rule simplifies complex polynomials into more manageable terms, streamlining the critical step of finding derivatives.
In the given exercise, the function \( u(x) = 2\sqrt{x} + 1 \) was first rewritten using the power format as \(2x^{1/2} + 1\). By applying the power rule, we differentiated \( 2x^{1/2} \) to get \( x^{-1/2} \), simplifying the computation significantly. As a practical and continuously applicable tool, mastering the power rule solidifies your calculus foundation, enabling you to tackle more complex equations with confidence.