Problem 87 - University Calculus: Early Transcendentals

Question

Suppose that functions \(f\) and \(g\) and their derivatives with respect to \(x\) have the following values at \(x=2\) and \(x=3\) $$\begin{array}{lcccc} \hline \boldsymbol{x} & \boldsymbol{f}(\boldsymbol{x}) & \boldsymbol{g}(\boldsymbol{x}) & \boldsymbol{f}^{\prime}(\boldsymbol{x}) & \boldsymbol{g}^{\prime}(\boldsymbol{x}) \\ \hline 2 & 8 & 2 & 1 / 3 & -3 \\ 3 & 3 & -4 & 2 \pi & 5 \\ \hline \end{array}$$ Find the derivatives with respect to \(x\) of the following combinations at the given value of \(x\) a. \(2 f(x), \quad x=2\) b. \(f(x)+g(x), \quad x=3\) c. \(f(x) \cdot g(x), \quad x=3\) d. \(f(x) / g(x), \quad x=2\) e. \(f(g(x)), \quad x=2\) f. \(\sqrt{f(x)}, \quad x=2\) g. \(1 / g^{2}(x), \quad x=3\) h. \(\sqrt{f^{2}(x)+g^{2}(x)}, \quad x=2\)

Step-by-Step Solution

Answer

(a) \(\frac{2}{3}\), (b) \(2\pi + 5\), (c) \(-8\pi + 15\), (d) \(\frac{37}{6}\), (e) \(-1\), (f) \(\frac{1}{12\sqrt{2}}\), (g) \(\frac{5}{16}\), (h) \(\frac{-5}{3\sqrt{68}}\).

Step 1: Derivative of Scalar Multiple

The derivative of a scalar multiple of a function is the scalar times the derivative of the function. For \(2f(x)\) when \(x=2\), the derivative is:\[\frac{d}{dx}[2f(x)] = 2f'(x)\]Given \(f'(2) = \frac{1}{3}\), the derivative is:\[2 \times \frac{1}{3} = \frac{2}{3}\]

Step 2: Derivative of a Sum

The derivative of a sum is the sum of the derivatives. For \(f(x) + g(x)\) at \(x=3\):\[\frac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x)\]Given \(f'(3) = 2\pi\) and \(g'(3) = 5\), the derivative is:\[2\pi + 5\]

Step 3: Product Rule

The product rule states that \((uv)' = u'v + uv'\). For \(f(x) \cdot g(x)\) at \(x=3\):\[\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)\]Using the values \(f(3) = 3\), \(g(3) = -4\), \(f'(3) = 2\pi\), \(g'(3) = 5\):\[(2\pi)(-4) + (3)(5) = -8\pi + 15\]

Step 4: Quotient Rule

The quotient rule states that \(\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}\). For \(\frac{f(x)}{g(x)}\) at \(x=2\):\[\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}\]With values \(f(2) = 8\), \(g(2) = 2\), \(f'(2) = \frac{1}{3}\), \(g'(2) = -3\):\[\frac{\left(\frac{1}{3}\cdot2\right) - (8)(-3)}{2^2} = \frac{\frac{2}{3} + 24}{4} = \frac{\frac{74}{3}}{4} = \frac{74}{12} = \frac{37}{6}\]

Step 5: Chain Rule

Using the chain rule for the composite function \(f(g(x))\), the derivative is \(f'(g(x))g'(x)\). At \(x=2\):\[\frac{d}{dx}[f(g(x))] = f'(g(2))g'(2)\]Using values \(g(2) = 2\), \(f'(2) = \frac{1}{3}\), and \(g'(2) = -3\):\[f'(2)g'(2) = \frac{1}{3} \cdot (-3) = -1\]

Step 6: Derivative of a Square Root Function

For the derivative of \(\sqrt{f(x)}\), use \(\frac{1}{2}f(x)^{-1/2} \cdot f'(x)\). At \(x=2\):\[\frac{d}{dx}[\sqrt{f(x)}] = \frac{1}{2\sqrt{f(x)}}f'(x)\]Using \(f(2) = 8\) and \(f'(2) = \frac{1}{3}\):\[\frac{1}{2\sqrt{8}} \cdot \frac{1}{3} = \frac{1}{2 \cdot 2\sqrt{2}} \cdot \frac{1}{3} = \frac{1}{12\sqrt{2}}\]

Step 7: Derivative of a Reciprocal Function

For the derivative of \(\frac{1}{g^2(x)}\), relate it to \(g(x)^{-2}\). The derivative is \(-2g(x)^{-3}g'(x)\). At \(x=3\):\[\frac{d}{dx}\left(\frac{1}{g^2(x)}\right) = -2g(x)^{-3}g'(x)\]Using \(g(3) = -4\) and \(g'(3) = 5\):\[-2(-4)^{-3}\cdot 5 = -2 \cdot \left(-\frac{1}{64}\right) \cdot 5 = \frac{10}{32} = \frac{5}{16}\]

Step 8: Derivative of a Composite Function (Distance Formula)

Use the chain rule for \(\sqrt{f^2(x) + g^2(x)}\) with derivative:\[\frac{1}{2\sqrt{f^2(x) + g^2(x)}}(2f(x)f'(x) + 2g(x)g'(x))\]At \(x=2\):\[\frac{2f(x)f'(x) + 2g(x)g'(x)}{2\sqrt{f^2(x) + g^2(x)}}\]Using \(f(2)=8\), \(g(2)=2\), \(f'(2)=\frac{1}{3}\), \(g'(2)=-3\):\[\frac{2(8)\frac{1}{3} + 2(2)(-3)}{2\sqrt{8^2 + 2^2}} = \frac{\frac{16}{3} - 12}{2\sqrt{68}} = \frac{\frac{-20}{3}}{2\cdot\sqrt{68}} = \frac{-20}{12\sqrt{68}} = \frac{-5}{3\sqrt{68}}\]

Key Concepts

Chain Rule

The Chain Rule is a fundamental concept for finding the derivative of composite functions, which are functions applied within other functions. When we look at a function like \( f(g(x)) \), it involves two layers. The function \( f \) is applied to the output of \( g \). To differentiate such functions, the chain rule tells us to multiply the derivative of the outer function by the derivative of the inner function.
For example, if \( u(x) = g(x) \) and \( v(u) = f(u) \), then \( y = f(g(x)) \) is \( v(u(x)) \). According to the chain rule:

• Find the derivative of \( v \) with respect to \( u \), which is \( v'(u) \).
• Next, find the derivative of \( u \) with respect to \( x \), which is \( u'(x) = g'(x) \).
• Finally, multiply these results: \( y'(x) = v'(u) \times g'(x) \).

This principle is essential when dealing with nested functions or when transforming a variable through several functions, ensuring a seamless process of differentiation.

Product Rule

In calculus, the Product Rule is indispensable when dealing with the derivative of the product of two functions. Suppose you have two functions \( u(x) \) and \( v(x) \). Their product, \( y = u(x)v(x) \), involves both multiplying the functions and finding their derivative. The Product Rule states:

• The derivative of \( y \) is \( y'(x) = u'(x)v(x) + u(x)v'(x) \).

Here’s the breakdown:
- Find \( u'(x) \), the derivative of \( u(x) \).
- Multiply it by \( v(x) \).
- Then, find \( v'(x) \), the derivative of \( v(x) \).
- Multiply it by \( u(x) \).
- The sum of these two products gives the derivative of the product.
This rule shines in problems involving multiplying functions, such as in physics where position and velocity functions multiply.

Quotient Rule

The Quotient Rule is a vital tool when differentiating a function that is expressed as a ratio of two other functions. If you have \( y = \frac{u(x)}{v(x)} \), the rule allows us to find \( y'(x) \). The Quotient Rule is given by:

• The derivative is \( y'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \).

Understanding each part is key:
- Compute \( u'(x) \) and multiply by \( v(x) \).
- Compute \( v'(x) \) and multiply by \( u(x) \).
- Subtract the second product from the first one.
- Divide this difference by \( [v(x)]^2 \), the square of the denominator.
This approach is especially useful in situations where the outputs are in fractions, helping to systematically tackle derivatives.

Composite Functions

Composite Functions, in calculus, involve operating one function within another. When you have \( f(g(x)) \), \( g(x) \) outputs a result, which then becomes the input for \( f \). Recognizing these compositions is crucial when determining derivatives using the chain rule. Notably, the chain rule simplifies this by applying:

• The derivative of the inner function \( g(x) \).
• Then multiply it by the derivative of the outer function,, evaluated at \( g(x) \).

The concept of composite functions extends beyond derivatives to other math areas. It's critical when transforming graphs, solving complex equations, or evaluating data, underpinning numerous real-life applications like computer science algorithms or physical transformations.