Problem 60 - Calculus Single Variable

Question

Given \(f(3)=-4\) and \(f^{\prime}(3)=6,\) find \((1 / f)^{\prime}(3)\) and \(\left(f^{2}\right)^{\prime}(3)\)

Step-by-Step Solution

Answer

\((1/f)'(3) = -\frac{3}{8}, (f^2)'(3) = -48\).

Step 1: Identify the Given Information

We are given that \(f(3) = -4\) and \(f'(3) = 6\). We need to find \((1 / f)'(3)\) and \((f^2)'(3)\).

Step 2: Find the Derivative of \(1/f\) Using the Quotient Rule

The function we are dealing with is \(g(x) = \frac{1}{f(x)}\), which is equivalent to \(g(x) = f(x)^{-1}\). The derivative, using the chain rule, is \(g'(x) = -f(x)^{-2} \cdot f'(x)\).

Step 3: Evaluate \((1/f)'(3)\)

Substitute \(f(3) = -4\) and \(f'(3) = 6\) into the derivative formula: \(\left(\frac{1}{f}\right)'(3) = -(-4)^{-2} \times 6 = -\frac{1}{16} \times 6 = -\frac{6}{16} = -\frac{3}{8}\).

Step 4: Calculate the Derivative of \(f^2\)

For \(h(x) = f(x)^2\), use the power rule combined with the chain rule: \(h'(x) = 2f(x) \cdot f'(x)\).

Step 5: Evaluate \((f^2)'(3)\)

Substitute \(f(3) = -4\) and \(f'(3) = 6\) into the derivative: \((f^2)'(3) = 2(-4) \times 6 = -8 \times 6 = -48\).

Key Concepts

Quotient Rule

When dealing with derivatives of functions represented as fractions, we use the quotient rule. This rule helps us differentiate a function of the form \(\frac{u(x)}{v(x)}\). The quotient rule states that:

• \(g'(x) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{(v(x))^2}\)

In our exercise, we used the function \(g(x) = \frac{1}{f(x)}\), which can be thought of as \(f(x)^{-1}\). To find the derivative using the quotient rule, we also utilize the chain rule by rewriting it as \(-f(x)^{-2} \cdot f'(x)\) to simplify the computation. This approach efficiently combines both rules to tackle more complex functions.
The quotient rule is particularly useful whenever a function is a ratio of two differentiable functions, making it easier to approach real-world problems.

Chain Rule

Understanding composite functions is made simple using the chain rule. This rule is used when you have a function within another function, such as \(f(g(x))\). The chain rule formula is:

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

This comes in handy when you want to differentiate complex expressions smoothly. In these scenarios, think of peeling an onion where each layer reveals a new derivative to compute.
In our problem, when finding the derivative of \(\frac{1}{f(x)}\) or \(f(x)^2\), we actually find ourselves using the chain rule to navigate through nested functions.
In essence, the chain rule allows us to break down intricate functions into simple parts, making differentiation a powerful tool for calculus problems.

Power Rule

The power rule is one of the most straightforward and widely used rules for differentiation. It allows us to take the derivative of functions where the variable is raised to a power. The rule states:

• If \(y = x^n\), then \(y' = n \cdot x^{n-1}\)

Utilizing this rule is essential for easy and quick computations when working with polynomial functions.
In the exercise given, we used the power rule to find the derivative of \(f(x) = f(x)^2\). The calculation involved not only this rule but also the chain rule, highlighting its utility in more complex scenarios.
Remember, by mastering the power rule, you unlock a gateway to solving more advanced differential calculus problems with just a bit more understanding and application.