Problem 32
Question
Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta\) as appropriate. $$y=\ln (\sec \theta+\tan \theta)$$
Step-by-Step Solution
Verified Answer
The derivative is \( \frac{\sec \theta \tan \theta + \sec^2 \theta}{\sec \theta + \tan \theta} \).
1Step 1: Identify the composition
The function given is \(y = \ln (\sec \theta + \tan \theta)\). The function is a composition of the natural logarithm and a trigonometric function inside it, which is \(\sec \theta + \tan \theta\). We use the chain rule to differentiate it.
2Step 2: Differentiate the outer function
Apply the chain rule and differentiate the outer function, which is the natural logarithm function \(\ln(u)\). The derivative of \(\ln(u)\) is \(\frac{1}{u}\). Substituting \(u = \sec \theta + \tan \theta\), we get \(\frac{1}{\sec \theta + \tan \theta}\).
3Step 3: Differentiate the inner function
Now, differentiate the inner function \(u = \sec \theta + \tan \theta\). The derivative of \(\sec \theta\) is \(\sec \theta \tan \theta\), and the derivative of \(\tan \theta\) is \(\sec^2 \theta\). Thus, the derivative of \(u\) is \(\sec \theta \tan \theta + \sec^2 \theta\).
4Step 4: Apply the chain rule
Using the chain rule, multiply the derivative of the outer function by the derivative of the inner function: \(\frac{1}{\sec \theta + \tan \theta} \times (\sec \theta \tan \theta + \sec^2 \theta)\).
5Step 5: Simplify the derivative expression
Combine the expressions: \[ \frac{1}{\sec \theta + \tan \theta} \times (\sec \theta \tan \theta + \sec^2 \theta) = \frac{\sec \theta \tan \theta + \sec^2 \theta}{\sec \theta + \tan \theta} \]. This fraction is the derivative of the original function.
Key Concepts
Trigonometric FunctionsNatural LogarithmDifferentiation
Trigonometric Functions
Trigonometric functions play a vital role in many areas of mathematics, especially in calculus. They include sine (\( \sin \theta \)), cosine (\( \cos \theta \)), and tangent (\( \tan \theta \)) among others. In this exercise, we focus on secant (\( \sec \theta \)) and tangent (\( \tan \theta \)).
Secant is the reciprocal of cosine, which means \( \sec \theta = \frac{1}{\cos \theta} \).
Tangent, on the other hand, is the sine of an angle divided by its cosine, or \( \tan \theta = \frac{\sin \theta}{\cos \theta} \).
Both secant and tangent functions are essential when solving differentiation problems involving trigonometric functions, as they each have unique derivatives. The derivative of \( \sec \theta \) is \( \sec \theta \tan \theta \), while the derivative of \( \tan \theta \) is \( \sec^2 \theta \).
These derivatives are useful when applying the chain rule, particularly in compositions of trigonometric functions, like in this exercise.
Secant is the reciprocal of cosine, which means \( \sec \theta = \frac{1}{\cos \theta} \).
Tangent, on the other hand, is the sine of an angle divided by its cosine, or \( \tan \theta = \frac{\sin \theta}{\cos \theta} \).
Both secant and tangent functions are essential when solving differentiation problems involving trigonometric functions, as they each have unique derivatives. The derivative of \( \sec \theta \) is \( \sec \theta \tan \theta \), while the derivative of \( \tan \theta \) is \( \sec^2 \theta \).
These derivatives are useful when applying the chain rule, particularly in compositions of trigonometric functions, like in this exercise.
Natural Logarithm
The natural logarithm, denoted as \( \ln(x) \), is a logarithm to the base of the mathematical constant \( e \), which is approximately 2.718.
It has unique properties that make it particularly important in calculus and complex calculations. One key property is that the derivative of \( \ln(x) \), with respect to \( x \), is \( \frac{1}{x} \).
In this specific problem, the function inside the logarithm is not purely \( x \), but a combination of trigonometric functions, namely \( \sec \theta + \tan \theta \).
Using the chain rule, we must differentiate both the outer natural logarithm function and the inner trigonometric function. This begins with recognizing that the outer function, \( \ln(u) \), contributes a derivative of \( \frac{1}{u} \) where \( u = \sec \theta + \tan \theta \).
This basic understanding of how the natural logarithm functions aids in unraveling more intricate expressions, like the one in our example.
It has unique properties that make it particularly important in calculus and complex calculations. One key property is that the derivative of \( \ln(x) \), with respect to \( x \), is \( \frac{1}{x} \).
In this specific problem, the function inside the logarithm is not purely \( x \), but a combination of trigonometric functions, namely \( \sec \theta + \tan \theta \).
Using the chain rule, we must differentiate both the outer natural logarithm function and the inner trigonometric function. This begins with recognizing that the outer function, \( \ln(u) \), contributes a derivative of \( \frac{1}{u} \) where \( u = \sec \theta + \tan \theta \).
This basic understanding of how the natural logarithm functions aids in unraveling more intricate expressions, like the one in our example.
Differentiation
Differentiation is a fundamental concept in calculus, referring to the process of finding the derivative of a function. The derivative represents the rate at which a function is changing at any point.
In problems like the one presented, dealing with compositions of functions, we often use the chain rule. The chain rule is a formula for calculating the derivative of a composition of two or more functions.
For example, if you have a composite function \( y = f(g(x)) \), then the derivative of \( y \) with respect to \( x \) can be found by \( f'(g(x)) \cdot g'(x) \). In context, this means we differentiate the outer function and multiply it by the derivative of the inner function.
In our problem, \( y = \ln(\sec \theta + \tan \theta) \), the outer function is \( \ln(x) \) and its derivative is \( \frac{1}{x} \). The inner function is \( \sec \theta + \tan \theta \), and its derivative, as calculated from the individual derivatives, is \( \sec \theta \tan \theta + \sec^2 \theta \).
Applying the chain rule, we have \( \frac{1}{\sec \theta + \tan \theta} \times (\sec \theta \tan \theta + \sec^2 \theta) \), which, upon simplification, yields the derivative of the composite function. By following these steps, students can gain a clear understanding of how differentiation, especially applying the chain rule, is crucial in calculus.
In problems like the one presented, dealing with compositions of functions, we often use the chain rule. The chain rule is a formula for calculating the derivative of a composition of two or more functions.
For example, if you have a composite function \( y = f(g(x)) \), then the derivative of \( y \) with respect to \( x \) can be found by \( f'(g(x)) \cdot g'(x) \). In context, this means we differentiate the outer function and multiply it by the derivative of the inner function.
In our problem, \( y = \ln(\sec \theta + \tan \theta) \), the outer function is \( \ln(x) \) and its derivative is \( \frac{1}{x} \). The inner function is \( \sec \theta + \tan \theta \), and its derivative, as calculated from the individual derivatives, is \( \sec \theta \tan \theta + \sec^2 \theta \).
Applying the chain rule, we have \( \frac{1}{\sec \theta + \tan \theta} \times (\sec \theta \tan \theta + \sec^2 \theta) \), which, upon simplification, yields the derivative of the composite function. By following these steps, students can gain a clear understanding of how differentiation, especially applying the chain rule, is crucial in calculus.
Other exercises in this chapter
Problem 32
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