Problem 62 - Calculus Single Variable

Question

Given \(f(4)=2, g(4)=1, f^{\prime}(4)=-5,\) and \(g^{\prime}(4)=-9,\) find \(\left(f \cdot g^{2}\right)^{\prime}(4),\left(g \cdot f^{2}\right)^{\prime}(4),\left(1 / f^{2}\right)^{\prime}(4),\) and \((g / f)^{\prime}(4)\)

Step-by-Step Solution

Answer

-41, -56, \(\frac{5}{4}\), \(-\frac{13}{4}\)

Step 1: Product Rule for First Function

We need to find \(\left(f \cdot g^2\right)'(4)\). Using the product rule, we express this as \(f'(4) \cdot g^2(4) + f(4) \cdot 2g(4)g'(4)\). Substitute the given values: \((-5) \cdot 1^2 + 2 \cdot 2 \cdot (-9)\). This evaluates to \(-5 - 36 = -41\).

Step 2: Product Rule for Second Function

Now, calculate \(\left(g \cdot f^2\right)'(4)\). Again, apply the product rule: \(g'(4) \cdot f^2(4) + g(4) \cdot 2f(4)f'(4)\). Substitute the known values: \((-9) \cdot 2^2 + 1 \cdot 2 \cdot (-5)\). The result is \(-36 - 20 = -56\).

Step 3: Power Rule and Chain Rule for Third Function

Calculate \(\left(\frac{1}{f^2}\right)'(4)\). Use the chain rule and rewrite the function as \(\left(f^2\right)'(4)^{-1}\). The derivative becomes \(-2\cdot f^{-3}(4) \cdot f'(4)\). Substitute the values: \(-2 \cdot (2)^{-3} \cdot (-5) = 10 \cdot \frac{1}{8} = \frac{5}{4}\).

Step 4: Quotient Rule for Fourth Function

Finally, determine \((g/f)'(4)\) using the quotient rule: \(\frac{f(4)g'(4) - g(4)f'(4)}{f^2(4)}\). With the values provided, this simplifies to \((2\cdot(-9) - 1\cdot(-5))/(2^2)\). Calculating gives \((-18 + 5)/4 = -13/4\).

Key Concepts

Product Rule

To dive into derivatives involving products of functions, we utilize the product rule. The product rule is a handy tool in calculus when differentiating products of two functions. For any two differentiable functions, say \( u(x) \) and \( v(x) \), the product rule states that the derivative of their product \( uv \) is given by:

• \( (uv)' = u'v + uv' \)

In the given problem, we first used this rule to find \( \left(f \cdot g^2\right)'(4) \). We express the derivative as:

• \( f'(4) \cdot g^2(4) + f(4) \cdot 2g(4)g'(4) \)

Substituting known values allows us to calculate easily. Understanding this approach helps in breaking down complex derivative expressions, making calculus problem solving more approachable and manageable.
Again, using the product rule for \( \left(g \cdot f^2\right)'(4) \), we approach with:

• \( g'(4) \cdot f^2(4) + g(4) \cdot 2f(4)f'(4) \)

Learning and practicing the product rule is crucial for any calculus student eager to tackle real-world problems that involve multiplying functions.

Quotient Rule

Calculating derivatives that involve division between two functions brings us to the quotient rule. The quotient rule is essential in calculus whenever you are dealing with derivatives of quotients of functions. Consider two functions \( u(x) \) and \( v(x) \) with \( v(x) eq 0 \); the derivative of \( \frac{u}{v} \) is given by:

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

In our problem, we utilized the quotient rule to find \( (g/f)'(4) \). Plugging in the values into our formula simplifies to:

• \( \frac{f(4)g'(4) - g(4)f'(4)}{f^2(4)} \)

With careful substitution, we calculated the outcome. Mastering the quotient rule allows you to handle any scenarios in calculus that involve dividing functions, whether in engineering, physics, or economics.

Chain Rule

The chain rule is a powerful technique for finding the derivative of composite functions. When you have a function within a function, the chain rule helps keep things simple. Suppose you have a composition of two functions \( f(g(x)) \), the derivative is found by:

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

In our example, to find \( \left(\frac{1}{f^2}\right)'(4) \), we first rewrite it using exponential notation as \( \left(f^2\right)^{-1} \). Applying the chain rule, we have:

• \(-2 \cdot f^{-3}(4) \cdot f'(4) \)

This approach simplifies solving complex derivatives, transforming intricate expressions into simpler ones by taking derivatives in steps using the appropriate rules. Grasping the chain rule equips you to unravel many real-world complexities in calculus.

Calculus Problem Solving

Calculus problem solving often involves applying multiple differentiation rules simultaneously. The key is identifying which rules, like the product, quotient, or chain rules, best fit the given functional expression. Through step-by-step analysis, decompose each part of the equation using their respective rules. This was evident in our exercise where different rules were applied for separate expressions.

• Identify the type of mathematical operation within the function (product, quotient, composite).
• Choose the appropriate rule (product rule for products, quotient rule for division, and chain rule for composites).
• Substitute known values to simplify and solve.

Skillful application of these principles is essential in higher mathematics and in fields requiring precise calculation techniques. Each rule has its place in calculus problem-solving, and knowing when and how to use them can make tackling derivatives much more straightforward and less intimidating.