Problem 28 - Calculus - AP Edition

Question

Find \(g^{\prime}(3)\) given that \(f(3)=-2\) and \(f^{\prime}(3)=4\). (a) \(g(x)=3 x^{2}-5 f(x)\) (b) \(g(x)=\frac{2 x+1}{f(x)}\)

Step-by-Step Solution

Answer

(a) \( g'(3) = -2 \); (b) \( g'(3) = -8 \).

Step 1: Differentiate part (a)

The function given for part (a) is \( g(x) = 3x^2 - 5f(x) \). To find \( g'(x) \), we need to differentiate each term separately. Differentiating \( 3x^2 \) gives \( 6x \). For the term \( -5f(x) \), by using the chain rule, we have \( -5f'(x) \). Therefore, \( g'(x) = 6x - 5f'(x) \).

Step 2: Substitute and solve for part (a)

Now, substitute \( x = 3 \), \( f(3) = -2 \), and \( f'(3) = 4 \) into the derivative. \( g'(3) = 6(3) - 5(4) = 18 - 20 = -2 \). Thus, \( g'(3) = -2 \).

Step 3: Differentiate part (b)

The function given for part (b) is \( g(x) = \frac{2x+1}{f(x)} \). Use the quotient rule, where \( u = 2x + 1 \) and \( v = f(x) \). The quotient rule states that \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \). So, \( u' = 2 \) and \( v' = f'(x) \). Applying the quotient rule, we get \( g'(x) = \frac{2f(x) - (2x+1)f'(x)}{f(x)^2} \).

Step 4: Substitute and solve for part (b)

Substitute \( x = 3 \), \( f(3) = -2 \), and \( f'(3) = 4 \) into the derivative. \( g'(3) = \frac{2(-2) - (2(3) + 1)(4)}{(-2)^2} = \frac{-4 - 28}{4} = \frac{-32}{4} = -8 \). Therefore, \( g'(3) = -8 \).

Key Concepts

Chain Rule

In calculus, the chain rule is an essential tool for differentiating composite functions. A composite function is one where a function is nested inside another. For instance, in our exercise, the expression \(-5f(x)\) involves a function \(f(x)\) within another function. To differentiate expressions like this, the chain rule allows us to handle each function layer one step at a time.

The chain rule states that if you have a composite function \(h(x) = g(f(x))\), where both \(g\) and \(f\) are differentiable, the derivative \(h'(x)\) can be found using the formula:

• \(h'(x) = g'(f(x)) \cdot f'(x)\)

This formula means that we first differentiate the outer function \(g\) with respect to \(f(x)\) and then multiply it by the derivative of \(f(x)\).

In step 1 of the original solution, the chain rule helps us differentiate the term \(-5f(x)\) by giving us \(-5f'(x)\), as we respect the inner function \(f(x)\) while considering how it changes.

Quotient Rule

The quotient rule is pivotal when you have a function that is the ratio of two differentiable functions. Think of the function as \(g(x) = \frac{u(x)}{v(x)}\). When it comes time to differentiate this, the quotient rule comes to the rescue.

The general formula of the quotient rule is as follows:

• \(\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}\)

What does this mean, exactly? It means that we differentiate the top \(u(x)\) and multiply it by the bottom \(v(x)\), then subtract the product of the top \(u(x)\) with the derivative of the bottom \(v(x)\).

In the original exercise, we have \(g(x) = \frac{2x+1}{f(x)}\). Applying the quotient rule here, \(u = 2x + 1\) and \(v = f(x)\). By following the formula to differentiate, we achieve the expression \(\frac{2f(x) - (2x+1)f'(x)}{f(x)^2}\). This systematically helps handle the more complex scenarios involving division.

Differentiation

Differentiation is a fundamental concept in calculus, dealing with how things change. At its core, differentiation tells us the rate of change of a function concerning one of its variables. It is akin to finding the slope of the function at any given point.

When differentiating, we apply various rules—like the chain rule and quotient rule—depending on the structure of the function. Each rule simplifies the process of finding derivatives, which are crucial for solving real-world problems, such as finding the speed, acceleration, or even optimizing a function.

In our context, when we differentiate expressions like \(3x^2\) in the exercise, we obtain \(6x\). Each term in a function has its derivative and, through the application of rules, gives insight into the behavior of complex expressions. Differentiation transforms into a toolbox, enabling us to understand and manipulate the dynamics of functions across multiple fields.