Question

Given \(r=e^{-p^{2}-q^{2}}, p=e^{s}, q=e^{-s},\) find \(d r / d s.\)

Short answer

\( \frac{dr}{ds} = e^{-(e^{2s} + e^{-2s})} \cdot (-2e^{2s} + 2e^{-2s}) \)

Step-by-step solution

Express the Given Variables

We need to express the variables in terms of s. Given that \( p = e^{s} \) and \( q = e^{-s} \) , substitute these into the expression for \( r \). So, \( r = e^{-p^{2} - q^{2}} \) becomes \( r = e^{-(e^{2s} + e^{-2s})} \).

Simplify the Expression for r

Rewrite the expression inside the exponent: \( r = e^{-(e^{2s} + e^{-2s})} \).

Differentiate r with respect to s

To find \( \frac{dr}{ds} \), differentiate both sides of \( r = e^{-(e^{2s} + e^{-2s})} \) with respect to s. \ Use the chain rule: \( \frac{dr}{ds} = e^{-(e^{2s} + e^{-2s})} \cdot \frac{d}{ds}[-(e^{2s} + e^{-2s})] \).

Differentiate the Exponent

Calculate \( \frac{d}{ds}[-(e^{2s} + e^{-2s})] \): \( \frac{d}{ds}[-e^{2s}] = -2e^{2s} \) and \( \frac{d}{ds}[-e^{-2s}] = 2e^{-2s} \). \ Therefore, \( \frac{d}{ds}[-(e^{2s} + e^{-2s})] = -2e^{2s} + 2e^{-2s} \).

Combine the Results

Now, combine the results from the previous steps: \( \frac{dr}{ds} = e^{-(e^{2s} + e^{-2s})} \cdot (-2e^{2s} + 2e^{-2s}) \).

Key concepts

Understanding the Chain Rule

The chain rule is a fundamental concept in calculus used to find the derivative of a composite function. If you have a function composed of two or more simpler functions, the chain rule allows you to break down the differentiation process.

For example, consider the function \(\r = f(g(x))\). To differentiate this using the chain rule, we first differentiate the outer function \(\f\) with respect to the inner function \(\f'\), and then multiply it by the derivative of the inner function \(g'(x)\). In symbolic form, the chain rule is written as:

\[ \frac{dr}{dx} = f'(g(x)) \times g'(x) \].

In our exercise, we used the chain rule to differentiate the function \(\r = e^{-(e^{2s} + e^{-2s})}\) by treating \(e^{2s} + e^{-2s}\) as the inner function. This step is crucial as it simplifies the differentiation process significantly

Applying the chain rule meant differentiating the exponent first, followed by multiplying it by the outer function's derivative.

Exploring Exponential Functions

Exponential functions are a key concept in calculus and appear frequently in various problems. They have the form \(\f(x) = e^{x}\), where \(e\) is the base of the natural logarithm, approximately equal to 2.718.

Some noteworthy properties of exponential functions include:

• The derivative of \(\f(x) = e^{x}\) with respect to \(x\) is itself, \(\f'(x) = e^{x}\).
• Exponential functions grow rapidly, making them useful for modeling real-world scenarios like population growth or radioactive decay.

In our exercise, we encounter exponential expressions like \(e^{2s}\) and \(e^{-2s}\). Differentiating these requires knowing basic exponential derivative rules:

• For \(\frac{d}{ds} e^{ks}\), the derivative is \(ke^{ks}\) where \(k\) is a constant. Thus, \( \frac{d}{ds} e^{2s} = 2e^{2s}\).
• For negative exponents, \( \frac{d}{ds} e^{-ks} = -ke^{-ks}\), so \( \frac{d}{ds} e^{-2s} = -2e^{-2s}\).

This knowledge directly applied when simplifying \( r \) and calculating \( \frac{dr}{ds} \).

Mastering Implicit Differentiation

Implicit differentiation is a technique for finding the derivative of a function when it’s not explicitly solved for one variable in terms of another. Instead, the function is given in terms of two or more variables simultaneously. This method is particularly useful for handling complex relationships between variables that aren’t easily separable.

To perform implicit differentiation, you:

• Differentiate both sides of the equation with respect to the independent variable while treating the dependent variable as a function of the independent variable.
• Apply the chain rule whenever necessary to account for the dependent variable’s derivative.

In our exercise, we didn't need to use implicit differentiation because we expressed the variables in terms of \(\r \) directly and simplified the exponent before differentiating. However, understanding implicit differentiation gives you a robust toolset for tackling more complex calculus problems, especially when direct differentiation isn’t feasible.

For more on implicit differentiation, consider the example of finding \( \frac{dy}{dx} \) for \(x^2 + y^2 = 1\).
Differentiate both sides with respect to \(x\): \[2x + 2y \frac{dy}{dx} = 0 \].
Solve for \( \frac{dy}{dx} \) to get: \[ \frac{dy}{dx} = -\frac{x}{y} \].
This showcases how implicit differentiation seamlessly handles derivatives when working with multiple variables.