If Y=tan U, U=v-1/v, And V=lnx, What Is The Value Of Dy/dx At X=e?
Understanding how to compute derivatives in complex composite functions is a fundamental skill in calculus, especially in the realms of advanced mathematics, engineering, and physics. When functions are nested within each other—such as in the case of \(Y = \tan U\), where \(U\) itself depends on \(V\), and \(V\) depends on \(X\)—it becomes essential to apply the chain rule meticulously. This problem exemplifies the application of multiple derivative rules combined, including the chain rule, product rule, and properties of logarithmic and trigonometric functions.
In this article, we will explore step-by-step how to find \(\frac{dy}{dx}\) given the composite functions:
- \(Y = \tan U\)
- \(U = v - \frac{1}{v}\)
- \(V = \ln x\)
and evaluate it specifically at \(x = e\). We will delve into the mathematical reasoning behind each step, providing detailed explanations to ensure clarity. This comprehensive guide aims to help students and enthusiasts understand the intricate process of differentiating nested functions and applying calculus principles efficiently.
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Understanding the Given Functions and Their Relationships
Before we proceed with differentiation, it’s crucial to understand the definitions and relationships among the functions involved:
- Function \(V = \ln x\): This function relates the independent variable \(x\) to an intermediate variable \(V\). The natural logarithm function \(\ln x\) is differentiable for \(x > 0\).
- Function \(U = v - \frac{1}{v}\): Here, \(U\) depends on \(V\), which in turn depends on \(x\). The function involves both a linear term and a reciprocal term of \(v\).
- Function \(Y = \tan U\): The output \(Y\) depends on \(U\), which depends on \(V\), and ultimately on \(x\).
Understanding these dependencies is key to applying the chain rule effectively. The goal is to compute \(\frac{dy}{dx}\) at the specific point where \(x = e\).
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Step-by-Step Derivation of \(\frac{dy}{dx}\)
The process involves multiple steps:
- Determine \(\frac{dy}{du}\): Derivative of \(Y = \tan U\) with respect to \(U\).
- Determine \(\frac{du}{dv}\): Derivative of \(U = v - \frac{1}{v}\) with respect to \(V\).
- Determine \(\frac{dv}{dx}\): Derivative of \(V = \ln x\) with respect to \(x\).
- Apply the chain rule: Combine these derivatives to find \(\frac{dy}{dx}\).
Let's explore each step in detail.
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1. Differentiating \(Y = \tan U\) with respect to \(U\)
The derivative of \(\tan U\) with respect to \(U\) is well-known:
\[
\frac{dy}{du} = \sec^2 U
\]
where \(\sec U = \frac{1}{\cos U}\). This derivative represents how \(Y\) changes with respect to the intermediate variable \(U\).
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2. Differentiating \(U = v - \frac{1}{v}\) with respect to \(V\)
Since \(U\) depends on \(V\) through \(v\), and \(v\) is a function of \(V\), the differentiation involves:
\[
\frac{du}{dv} = 1 - \frac{d}{dv}\left(\frac{1}{v}\right)
\]
The derivative of \(1/v\) with respect to \(v\) is:
\[
\frac{d}{dv}\left(\frac{1}{v}\right) = -\frac{1}{v^2}
\]
Thus,
\[
\frac{du}{dv} = 1 - \left(-\frac{1}{v^2}\right) = 1 + \frac{1}{v^2}
\]
This derivative indicates how \(U\) varies with \(V\).
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3. Differentiating \(V = \ln x\) with respect to \(x\)
The derivative of \(V = \ln x\) with respect to \(x\) is straightforward:
\[
\frac{dv}{dx} = \frac{1}{x}
\]
This expresses the rate of change of the logarithmic function relative to \(x\).
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4. Combining derivatives using the chain rule
Applying the chain rule to find \(\frac{dy}{dx}\):
\[
\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dv} \times \frac{dv}{dx}
\]
Substituting the derivatives obtained:
\[
\frac{dy}{dx} = \sec^2 U \times \left(1 + \frac{1}{v^2}\right) \times \frac{1}{x}
\]
To evaluate \(\frac{dy}{dx}\) at \(x = e\), we need to compute \(U\) and \(v\) at \(x = e\).
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Evaluating the Derivative at \(x = e\)
Let's now compute the necessary values at \(x = e\).
Step 1: Find \(V\) at \(x = e\)
\[
V = \ln x \Rightarrow V = \ln e = 1
\]
Step 2: Find \(v\) at \(V = 1\)
Since \(V = \ln x\), and \(v = V\), we have:
\[
v = 1
\]
Step 3: Find \(U\) at \(v = 1\)
\[
U = v - \frac{1}{v} = 1 - \frac{1}{1} = 1 - 1 = 0
\]
Step 4: Compute the derivatives at these values
- \(\sec^2 U\) at \(U=0\):
\[
\sec^2 0 = \left(\frac{1}{\cos 0}\right)^2 = 1^2 = 1
\]
- \(1 + \frac{1}{v^2}\) at \(v=1\):
\[
1 + \frac{1}{1^2} = 1 + 1 = 2
\]
- \(x = e\):
\[
\frac{1}{x} = \frac{1}{e}
\]
Putting it all together:
\[
\left.\frac{dy}{dx}\right|_{x=e} = 1 \times 2 \times \frac{1}{e} = \frac{2}{e}
\]
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Final Answer and Summary
The value of \(\frac{dy}{dx}\) at \(x = e\) for the given nested functions is:
\[
\boxed{\frac{2}{e}}
\]
This result encapsulates the interplay of logarithmic, reciprocal, and trigonometric functions within the calculus framework, demonstrating the power of the chain rule in handling complex derivatives.
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Additional Insights and Applications
Calculating derivatives of composite functions such as these is not only an academic exercise but also vital in various real-world applications. Here are some contexts where understanding such derivatives is important:
- Physics: Modeling systems where the rate of change depends on multiple interdependent variables, such as in thermodynamics or quantum mechanics.
- Engineering: Signal processing and control systems often involve nested functions, requiring precise derivative calculations to analyze system stability.
- Economics: Marginal analysis where quantities depend on multiple nested functions, like utility functions or cost functions.
- Mathematics Education: Developing problem-solving skills and deepening understanding of calculus principles.
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Conclusion
Mastering the differentiation of nested functions is a fundamental aspect of advanced calculus. By carefully applying the chain rule and understanding the relationships among functions, complex derivatives become manageable. In the specific problem tackled here, we have systematically broken down each component, evaluated at the point of interest, and obtained a precise numerical result.
Remember, practice is key to mastering these techniques. Attempting similar problems with different functions will strengthen your calculus skills and prepare you for more challenging mathematical scenarios. Whether you're a student, a teacher, or a professional, a solid grasp of derivatives in complex functions is an invaluable tool in your mathematical toolkit.
