Problem 14 - Calculus Early Transcendentals: Pearson New International Edition

Question

\(G(v)=\frac{\sin v \cos v}{\sqrt{1+\cos ^{2} v}}, \quad[0, \pi / 2]\)

Step-by-Step Solution

Answer

Differentiate using the product and quotient rules, simplify using trigonometric identities.

Step 1: Differentiate the Numerator

The numerator of the function is \( \sin v \cos v \). We will use the product rule for differentiation, which states \( \frac{d}{dv}[u \, v] = u' \, v + u \, v' \), where \( u = \sin v \) and \( v = \cos v \). The derivatives are \( u' = \cos v \) and \( v' = -\sin v \). Hence, the derivative of the numerator is:\[ \frac{d}{dv}(\sin v \cos v) = \cos v \cdot \cos v + \sin v \cdot (-\sin v) = \cos^2 v - \sin^2 v. \]

Step 2: Differentiate the Denominator

The denominator of the function is \( \sqrt{1 + \cos^2 v} \). First, find the derivative of \( 1 + \cos^2 v \), which is \( -2\cos v \sin v \). Then use the chain rule to differentiate \( \sqrt{1 + \cos^2 v} \), resulting in:\[ \frac{d}{dv}[\sqrt{1 + \cos^2 v}] = \frac{1}{2\sqrt{1 + \cos^2 v}} \cdot (-2\cos v \sin v) = \frac{-\cos v \sin v}{\sqrt{1 + \cos^2 v}}. \]

Step 3: Use the Quotient Rule

Apply the quotient rule \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \), where \( u = \sin v \cos v \) and \( v = \sqrt{1 + \cos^2 v} \). Substitute the derivatives:\[G'(v) = \frac{(\cos^2 v - \sin^2 v) \sqrt{1 + \cos^2 v} - (\sin v \cos v) \frac{-\cos v \sin v}{\sqrt{1 + \cos^2 v}}}{1 + \cos^2 v}.\]

Step 4: Simplify the Expression

Combining and simplifying the expressions from the quotient rule, we get:\[G'(v) = \frac{\cos^2 v - \sin^2 v + \frac{(\sin^2 v \cos^2 v)}{1 + \cos^2 v}}{1 + \cos^2 v}\]Continue simplifying by collecting like terms and analyzing how the components interact. However, direct simplification requires considering trigonometric identities and algebraic manipulation, achieving a textbook derivative solves the differential change.

Key Concepts

Product Rule

In calculus, when differentiating a product of two functions, we employ the product rule. This concept is fundamental for handling expressions like \( f(x) = u(x) \, v(x) \). The product rule formula is:

• \( \frac{d}{dx}[u(x) \, v(x)] = u'(x) \, v(x) + u(x) \, v'(x) \)

Here, \( u(x) \) and \( v(x) \) are differentiable functions. The process involves differentiating each function separately, then combining the results using multiplication and addition. For example, to differentiate \( \sin v \cos v \) in the exercise, we set \( u = \sin v \) and \( v = \cos v \) resulting in \( u' = \cos v \) and \( v' = -\sin v \). Thus:

• \( \frac{d}{dv}(\sin v \cos v) = \cos v \cdot \cos v + \sin v \cdot (-\sin v) = \cos^2 v - \sin^2 v \)

Mastering the product rule is crucial as it comes up across various problems, particularly when dealing with products of trigonometric or polynomial functions.

Quotient Rule

The quotient rule is used to differentiate a division of two functions, often appearing in expressions like \( \frac{u(x)}{v(x)} \). The rule can be a bit more complex but follows a similar structured approach like the product rule. The formula is:

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

First, differentiate the numerator to find \( u' \) and the denominator to find \( v' \). In the exercise above, \( u(x) = \sin v \cos v \) and \( v(x) = \sqrt{1 + \cos^2 v} \), leading to their respective derivatives found in previous steps. When applying the quotient rule, it's vital to keep the order of subtraction correct in the numerator to avoid sign errors. Simplification is typically necessary to reach the final result, as seen in the original solution. Simplifying afterwards will let you clean up complex results, revealing the derivative's core traits.

Chain Rule

The chain rule helps differentiate composite functions, those made by wrapping one function inside another. It is particularly useful when encountering square roots, powers, or combinations, like \( \sqrt{1 + \cos^2 v} \) in the exercise. The chain rule is expressed as:

• \( \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \)

In essence, differentiate the outer function while keeping the inner function unchanged, then multiply by the derivative of the inner function. For the example \( \sqrt{1+ \cos^2 v} \), consider it as \( (1 + \cos^2 v)^{1/2} \). First, find the derivative of \( 1 + \cos^2 v \), which is \( -2\cos v \sin v \), then multiply by the derivative of the square root function.

• Chain rule application yields \( \frac{1}{2\sqrt{1 + \cos^2 v}} \cdot (-2\cos v \sin v) \)
• Totaling to: \( \frac{-\cos v \sin v}{\sqrt{1 + \cos^2 v}} \)

Through practice, applying the chain rule can become an automatic process, essential for tackling complex differentiation problems.

Trigonometric Derivatives

Trigonometric functions are everywhere in calculus, and knowing their derivatives is essential. These derivatives stem from fundamental trigonometric identities and behavior:

• The derivative of \( \sin x \) is \( \cos x \)
• The derivative of \( \cos x \) is \( -\sin x \)

These basic derivatives form the backbone for differentiating more complex trigonometric expressions. For instance, in our exercise, both \( \sin v \) and \( \cos v \) derivatives are essential for both the product and chain rules. Understanding these base derivatives allows you to predict how functions behave and helps identifying patterns in exercises. Furthermore, employing identities like \( \cos^2 v - \sin^2 v \) can be particularly useful in simplifying results or matching them with known trigonometric identities. Mastery of trigonometric derivatives not only facilitates solving calculus problems but enhances comprehension of the interplay between different components in complex functions.