Chapter 9: Problem 49
Use (8) to find the indicated derivative. $$ z=\ln \left(u^{2}+v^{2}\right) ; u=t^{2}, v=t^{-2} ; \frac{d z}{d t} $$
Short Answer
Expert verified
\( \frac{dz}{dt} = \frac{4t^3 - 4t^{-5}}{t^4 + t^{-4}} \).
Step by step solution
01
Identify the Function and Substitutions
We have the function \( z = \ln(u^2 + v^2) \) with substitutions \( u = t^2 \) and \( v = t^{-2} \). Our goal is to find the derivative \( \frac{dz}{dt} \).
02
Substitute u and v into the Function
Substitute the expressions for \( u \) and \( v \) into \( z \). This gives \( z = \ln((t^2)^2 + (t^{-2})^2) = \ln(t^4 + t^{-4}) \).
03
Differentiate the Function with Respect to t
To find \( \frac{dz}{dt} \), apply the chain rule. The derivative of \( \ln(f(t)) \) is \( \frac{1}{f(t)} \cdot \frac{df(t)}{dt} \), where \( f(t) = t^4 + t^{-4} \).
04
Compute the Derivative of f(t)
Differentiate \( f(t) = t^4 + t^{-4} \) with respect to \( t \). The derivative is \( \frac{d}{dt}(t^4 + t^{-4}) = 4t^3 - 4t^{-5} \).
05
Combine Using the Chain Rule
Using the chain rule, substitute back: \( \frac{dz}{dt} = \frac{1}{t^4 + t^{-4}} \cdot (4t^3 - 4t^{-5}) \).
06
Simplify the Expression
Simplify the expression \( \frac{dz}{dt} = \frac{4t^3 - 4t^{-5}}{t^4 + t^{-4}} \) to its simplest form, if possible.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chain Rule
The chain rule is a fundamental tool in calculus used to differentiate composite functions. It allows us to find the derivative of a function that has another function inside it. If a function can be expressed as a composition of two functions, such as \( z = g(f(t)) \), the derivative \( \frac{dz}{dt} \) is found by multiplying the derivative of the outer function \( g \) evaluated at \( f(t) \) by the derivative of the inner function \( f(t) \).
For example, in the provided exercise, we have \( z = \ln(u^2 + v^2) \) where \( u=t^2 \) and \( v=t^{-2} \). To find \( \frac{dz}{dt} \), we substitute the expressions for \( u \) and \( v \) into \( f(t) \) to get \( f(t) = t^4 + t^{-4} \). The chain rule is then applied by first differentiating \( \ln(f(t)) \), which results in \( \frac{1}{f(t)} \) times the derivative of \( f(t) \) with respect to \( t \). This method breaks down complex functions into manageable steps and simplifies the differentiation process.
For example, in the provided exercise, we have \( z = \ln(u^2 + v^2) \) where \( u=t^2 \) and \( v=t^{-2} \). To find \( \frac{dz}{dt} \), we substitute the expressions for \( u \) and \( v \) into \( f(t) \) to get \( f(t) = t^4 + t^{-4} \). The chain rule is then applied by first differentiating \( \ln(f(t)) \), which results in \( \frac{1}{f(t)} \) times the derivative of \( f(t) \) with respect to \( t \). This method breaks down complex functions into manageable steps and simplifies the differentiation process.
- Differentiate the outer function.
- Multiply by the derivative of the inner function.
- Simplify the resulting expression if possible.
Natural Logarithm Differentiation
Differentiating a function that involves a natural logarithm is straightforward once you understand the basic rule: the derivative of \( \ln(x) \) is \( \frac{1}{x} \). When dealing with a compound function like \( \ln(u^2 + v^2) \), where \( u \) and \( v \) are functions of another variable \( t \), the process involves both the natural logarithm rule and the chain rule.
In this exercise, the function inside the logarithm is \( u^2 + v^2 \), substituted with \( t^4 + t^{-4} \) in terms of \( t \). Hence, the derivative of \( z = \ln(t^4 + t^{-4}) \) gives \( \frac{1}{t^4 + t^{-4}} \). The differentiation not only requires applying the logarithmic derivative rule but also calls for the subsequent differentiation of \( t^4 + t^{-4} \) itself.
This process highlights how these rules combine to tackle more complex expressions:
In this exercise, the function inside the logarithm is \( u^2 + v^2 \), substituted with \( t^4 + t^{-4} \) in terms of \( t \). Hence, the derivative of \( z = \ln(t^4 + t^{-4}) \) gives \( \frac{1}{t^4 + t^{-4}} \). The differentiation not only requires applying the logarithmic derivative rule but also calls for the subsequent differentiation of \( t^4 + t^{-4} \) itself.
This process highlights how these rules combine to tackle more complex expressions:
- Apply the logarithm differentiation rule.
- Use the result to multiply by the derivative of the internal function.
- Simplify where possible for clarity.
Mathematical Substitution
Mathematical substitution is a technique often used to simplify differentiation tasks by expressing a given function in terms of new variables. This approach enables a clearer pathway to applying differentiation rules, especially in more complex multi-variable scenarios.
In the exercise, substitutions like \( u = t^2 \) and \( v = t^{-2} \) replace the variables in the function \( z = \ln(u^2 + v^2) \). By doing so, we convert the function into a single variable function \( f(t) = t^4 + t^{-4} \), which is easier to differentiate. Substitution reduces the complexity by focusing on a straightforward variable representation.
Here’s a simplified approach to perform substitutions effectively:
In the exercise, substitutions like \( u = t^2 \) and \( v = t^{-2} \) replace the variables in the function \( z = \ln(u^2 + v^2) \). By doing so, we convert the function into a single variable function \( f(t) = t^4 + t^{-4} \), which is easier to differentiate. Substitution reduces the complexity by focusing on a straightforward variable representation.
Here’s a simplified approach to perform substitutions effectively:
- Identify substitute expressions for complicated parts of the function.
- Re-express the function using these new terms.
- Differentiation becomes manageable and often, standard rule applications follow.
- Substitute back only if necessary to interpret results.