Question

Using Relationships Given that \(g(5)=-3, g^{\prime}(5)=6,\) \(h(5)=3,\) and \(h^{\prime}(5)=-2,\) find \(f^{\prime}(5)\) for each of the following, if possible. If it is not possible, state what additional information is required. $$ \begin{array}{ll}{\text { (a) } f(x)=g(x) h(x)} & {\text { (b) } f(x)=g(h(x))} \\\ {\text { (c) } f(x)=\frac{g(x)}{h(x)}} & {\text { (d) } f(x)=[g(x)]^{3}}\end{array} $$

Short answer

The computed values of \(f'(5)\) for the functions are A) 24, B) -12, C) 2, D) 162.

Step-by-step solution

A) Apply the Product Rule

Given \(f(x)=g(x) h(x)\), we need to apply the product rule for differentiation. It states that \( (u \cdot v)' = u' \cdot v + u \cdot v' \). So, \(f'(x) = g'(x)h(x) + g(x)h'(x)\). Substituting the given values for \(g(5), g'(5), h(5)\) and \(h'(5)\), we get: \(f'(5) = 6*3 + (-3)*(-2) = 24\)

B) Apply the Chain Rule

Given \(f(x)=g(h(x))\), apply chain rule: \( (g \circ h)'(x) = g'(h(x)) \cdot h'(x)\). Substituting the given values, we get: \(f'(5) = 6*(-2) = -12\).

C) Apply the Quotient Rule

Given \(f(x)=\frac{g(x)}{h(x)}\), apply quotient rule: \((\frac{u}{v})' = \frac{u'v - uv'}{v^2}\). Substituting the given values, we get: \(f'(5) = \frac{6*3 - (-3)*(-2)}{3^2} = 2\).

D) Apply the Chain Rule Again

Given \(f(x)=[g(x)]^{3}\), use chain rule: \((u^n)' = n \cdot u' \cdot u^{(n-1)}\). Hence, \(f'(x) = 3[g'(x)] \cdot [g(x)]^{2}\). Substituting the given values, we get: \(f'(5) = 3*6*(-3)^2 = 162\).

Key concepts

Product Rule in Calculus

The product rule is a fundamental operation used when differentiating products of two functions, symbolically represented as when we have a function of the form, \(f(x) = u(x) \cdot v(x)\). According to the product rule, the derivative of this product is \(f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)\)

In the context of our exercise, where \(f(x)=g(x)h(x)\), using the given values, the application of the product rule simplifies to \(f'(5) = g'(5)h(5) + g(5)h'(5)\). After substituting the respective values, it's clear that understanding the product rule equips you with the ability to solve such problems where two functions are being multiplied.

Chain Rule for Function Composition

The chain rule is a powerful and widely used technique in calculus for differentiating compositions of functions. It expresses the derivative of a composite function in terms of the derivatives of its inner and outer functions. Mathematically, if \(h(x)\) is a function and \(g(x)\) is another function, for the composite function \(f(x) = g(h(x))\), the derivative is \(f'(x) = g'(h(x)) \cdot h'(x)\).

In our exercise, for parts (b) and (d), the chain rule is utilized for different scenarios. For part (b), where the function \(f(x)\) is directly the composition of \(g\) and \(h\), we use the rule straightforwardly. In part (d), however, we deal with \(g(x)\) to the power of 3, which is also treated as a chain of functions—first raising \(g(x)\) to a power, then multiplying by its derivative. The beauty of the chain rule is in its ability to simplify what appears to be a complex differentiation problem into manageable parts.

Quotient Rule for Differentiating Ratios

The quotient rule is another essential derivative rule used when a function is represented as the ratio of two other functions, essentially \(f(x)=\frac{u(x)}{v(x)}\). The rule states that the derivative of this function is \(f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2}\).

Applied to our exercise, part (c), we see the quotient rule in action. With \(f(x)=\frac{g(x)}{h(x)}\), and after substituting in the known values for \(g(5)\), \(g'(5)\), \(h(5)\), and \(h'(5)\), the quotient is differentiated with ease. The quotient rule is particularly useful because it structures the way in which we find the derivative of a division, ensuring precision and reducing the possibility of error.

Understanding Function Differentiation

Function differentiation, at its core, is about finding the derivative of a function, which represents the rate at which the function is changing at any given point. It's a fundamental concept in calculus, critical for modeling and analyzing dynamic systems.

Differentiation rules, like the product, chain, and quotient rules, offer structured methods for tackling complex differential tasks. The exercise showcases the practical use of these rules, making clear how differentiation acts as a powerful tool in understanding the behavior of mathematical functions. By mastering these rules, students can take abstract functions and reveal their real-world implications — from understanding rates of change in physics to optimizing revenue in business.