Problem 19
Question
Find the derivative of \(y\) with respect to the given independent variable. \(y=7^{\sec \theta} \ln 7\)
Step-by-Step Solution
Verified Answer
\(\frac{dy}{d\theta} = \ln 7 \cdot 7^{\sec \theta} \sec \theta \tan \theta\).
1Step 1: Identify Function Composition
The given function is a composition of two functions: an exponential function and a logarithmic function: \( y = 7^{\sec \theta} \ln 7 \). The outer function is the logarithmic function with respect to the base \(7\), and the inner function is \(\sec \theta\) which serves as the exponent.
2Step 2: Apply the Chain Rule
To differentiate a composition of functions, we need to apply the chain rule. Identify \( u = \sec \theta \) such that \( y = 7^{u} \ln 7 = \ln 7 \cdot 7^{\sec \theta}\). The derivative of \(7^u\) with respect to \(u\) is \(\ln 7 \cdot 7^u\). We must multiply by the derivative of \(u = \sec \theta\) with respect to \(\theta\).
3Step 3: Differentiate Inner Function
Differentiate \(u = \sec \theta\) with respect to \(\theta\). The derivative is \(\frac{du}{d\theta} = \sec \theta \tan \theta\).
4Step 4: Apply the Chain Rule Fully
Now, apply the chain rule: \[ \frac{dy}{d\theta} = \ln 7 \cdot 7^{\sec \theta} \cdot \sec \theta \tan \theta \].
5Step 5: Simplify the Derivative
The final expression for the derivative is simplified as: \[ \frac{dy}{d\theta} = \ln 7 \cdot 7^{\sec \theta} \sec \theta \tan \theta \].
Key Concepts
Chain RuleExponential FunctionLogarithmic Function
Chain Rule
The chain rule is a fundamental tool in calculus used to find the derivative of composite functions. These are functions nested within functions, often seen in complex expressions. When faced with differentiating such functions, the chain rule simplifies the process by allowing us to break it down into manageable parts.
Here's a simple way to think about it:
In the original exercise, the function is \(y = 7^{\sec \theta} \ln 7\). Here, \(f(u) = 7^u\) and \(g(\theta) = \sec \theta\), with \(u = \sec \theta\). The chain rule helps us differentiate this efficiently by multiplying the derivative of the outer function \(7^u\) by the derivative of the inner function \(\sec \theta\).
Here's a simple way to think about it:
- Identify the outer function and the inner function. The outer function acts on the result of the inner function.
- Differentiate the outer function with respect to the inner one.
- Multiply this result by the derivative of the inner function.
In the original exercise, the function is \(y = 7^{\sec \theta} \ln 7\). Here, \(f(u) = 7^u\) and \(g(\theta) = \sec \theta\), with \(u = \sec \theta\). The chain rule helps us differentiate this efficiently by multiplying the derivative of the outer function \(7^u\) by the derivative of the inner function \(\sec \theta\).
Exponential Function
Exponential functions are a vital part of mathematics, characterized by constant rates of growth. Specifically, functions of the form \(a^x\), where \(a\) is a positive constant, are particularly significant. These functions have important properties, like the base \(e\), which naturally rises out of calculus.
In the problem provided, we have an exponential function \(7^{\sec \theta}\). A unique component to our function is the variable exponent \(\sec \theta\). This makes its differentiation a practical example of the chain rule applied to exponential functions.
For exponential functions of the form \(a^u\), where \(u\) is an expression in terms of a variable, the derivative is:\[ \frac{d}{dx}[a^u] = \ln(a) \cdot a^u \cdot \frac{du}{dx} \]
This formula is used in the solution to calculate \(\frac{dy}{d\theta}\), where the exponential function \(7^{\sec \theta}\) is differentiated to give \(\ln 7 \cdot 7^{\sec \theta} \cdot \frac{du}{d\theta}\). This highlights the multiplier role of the logarithm of the base in exponential differentiation.
In the problem provided, we have an exponential function \(7^{\sec \theta}\). A unique component to our function is the variable exponent \(\sec \theta\). This makes its differentiation a practical example of the chain rule applied to exponential functions.
For exponential functions of the form \(a^u\), where \(u\) is an expression in terms of a variable, the derivative is:\[ \frac{d}{dx}[a^u] = \ln(a) \cdot a^u \cdot \frac{du}{dx} \]
This formula is used in the solution to calculate \(\frac{dy}{d\theta}\), where the exponential function \(7^{\sec \theta}\) is differentiated to give \(\ln 7 \cdot 7^{\sec \theta} \cdot \frac{du}{d\theta}\). This highlights the multiplier role of the logarithm of the base in exponential differentiation.
Logarithmic Function
Logarithmic functions are the inverse of exponential functions. They help solve for the time or rate necessary to reach a certain level of growth. A key property of logarithms is that they transform multiplication into addition, simplifying many algebraic equations and derivatives.
In mathematics, the natural logarithm, \(\ln\), is most frequently used with the base \(e\). However, in the current problem, we deal with a logarithm of base 7 due to the constant \(7^{\sec \theta}\). Although it appears as a constant multiplier in this context, its role is crucial because when differentiating expressions like \(7^u\), the base's logarithm (\(\ln 7\)) is involved in the derivative formula.
Thus, having \(y = 7^{\sec \theta} \ln 7\) means this natural logarithm serves to scale the resulting derivative of the base power. Understanding this helps in working with both sides of the exponential-logarithmic relationship and fully grasping the role logarithms play in deriving these functions efficiently.
In mathematics, the natural logarithm, \(\ln\), is most frequently used with the base \(e\). However, in the current problem, we deal with a logarithm of base 7 due to the constant \(7^{\sec \theta}\). Although it appears as a constant multiplier in this context, its role is crucial because when differentiating expressions like \(7^u\), the base's logarithm (\(\ln 7\)) is involved in the derivative formula.
Thus, having \(y = 7^{\sec \theta} \ln 7\) means this natural logarithm serves to scale the resulting derivative of the base power. Understanding this helps in working with both sides of the exponential-logarithmic relationship and fully grasping the role logarithms play in deriving these functions efficiently.
Other exercises in this chapter
Problem 19
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