Question

If \(f(3)=7, f^{\prime}(3)=2, g(3)=6\), and \(g^{\prime}(3)=-10\), find (a) \((f-g)^{\prime}(3)\) (b) \((f \cdot g)^{\prime}(3)\) (c) \((g / f)^{\prime}(3)\)

Short answer

(a) 12, (b) -58, (c) \(-\frac{82}{49}\).

Step-by-step solution

Understanding the problem

We are given the values of functions and their derivatives at a certain point. Specifically, \(f(3) = 7\), \(f'(3) = 2\), \(g(3) = 6\), and \(g'(3) = -10\). We need to find the derivatives of the functions \(f-g\), \(f \cdot g\), and \(g / f\) at \(x = 3\).

Calculate \((f-g)'(3)\)

The derivative of the difference of two functions is the difference of their derivatives. So, \((f-g)'(x) = f'(x) - g'(x)\). Substituting \(x = 3\), we have:\[(f-g)'(3) = f'(3) - g'(3) = 2 - (-10) = 2 + 10 = 12.\]

Calculate \((f \cdot g)'(3)\)

Use the product rule for differentiation: \((f \cdot g)'(x) = f'(x)g(x) + f(x)g'(x)\).Substituting \(x = 3\), we get:\[(f \cdot g)'(3) = f'(3)g(3) + f(3)g'(3) = 2 \cdot 6 + 7 \cdot (-10) = 12 - 70 = -58.\]

Calculate \((g / f)'(3)\)

Utilize the quotient rule for differentiation: \((\frac{g}{f})'(x) = \frac{f(x)g'(x) - g(x)f'(x)}{(f(x))^2}\).Substituting \(x = 3\), we have:\[(\frac{g}{f})'(3) = \frac{7 \cdot (-10) - 6 \cdot 2}{7^2} = \frac{-70 - 12}{49} = \frac{-82}{49}.\]

Key concepts

Product Rule

The product rule is an essential tool in calculus for finding the derivative of the product of two functions. When you have two functions, say \( u(x) \) and \( v(x) \), their product is \( u(x) \, v(x) \). To find the derivative of this product, you use the product rule. It states:

• \( (u \, v)'(x) = u'(x) \, v(x) + u(x) \, v'(x) \)

This means you take the derivative of the first function and multiply it by the second function, then add the product of the first function and the derivative of the second function. This rule is crucial because it allows us to handle products of functions without directly multiplying them out first.
In our problem, we're applying the product rule to find \((f \cdot g)'(3)\). By substituting the given values, we not only practice the application but also see how powerful the product rule is in real scenarios.

Quotient Rule

The quotient rule is similar to the product rule but is used when two functions are divided by each other. If you have two functions, \( u(x) \) over \( v(x) \), the quotient rule helps find the derivative of \( \frac{u(x)}{v(x)} \). The rule is expressed as:

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

This equation might seem a bit complicated at first, but it follows a consistent pattern. You take the derivative of the numerator, multiply it by the denominator, subtract the product of the numerator and the derivative of the denominator, and then divide by the square of the denominator. In our exercise, we're using this rule to find \((\frac{g}{f})'(3)\), which emphasizes the elegance and utility of the quotient rule in simplifying complex derivatives.

Function Derivatives

Understanding the concept of function derivatives is foundational in calculus. A derivative essentially measures how a function changes as its input changes. For a given function \( f(x) \), the derivative \( f'(x) \) represents the rate of change or the slope of the function at any point \( x \).
The process of finding this derivative is known as differentiation. Derivatives are everywhere in calculus and are used to find tangents, optimize solutions, and even model physical systems like motion. In our example, we're given the derivatives \( f'(3) = 2 \) and \( g'(3) = -10 \), which tells us how the graphs of these functions are behaving at \( x = 3 \). By working with these values, we explore the immediate rate of change of these functions.

Differentiation Techniques

Differentiation techniques are methods used to find the derivative of functions, which can range from simple polynomials to complex compositions. Some key techniques include:

• Power Rule: \( \frac{d}{dx} [x^n] = nx^{n-1} \)
• Sum/Difference Rule: \( (u \pm v)'(x) = u'(x) \pm v'(x) \)
• Product Rule and Quotient Rule (as discussed earlier)
• Chain Rule: Used for the composition of functions, \( (f(g(x)))' = f'(g(x))g'(x) \)

These techniques are integral to solving calculus problems efficiently. By combining these rules, we can tackle almost any differentiation problem, breaking down complex expressions into manageable parts. In the exercise, these techniques help us systematically find the derivatives of function combinations like \( f-g \), \( f \cdot g \), and \( g/f \). Each technique serves as a building block in our mathematical toolkit.