Problem 13 - Calculus with Concepts in Calculus

Question

Compute \(\partial z / \partial r\) and \(\partial z / \partial s\). $$ \mathrm{z}=u e^{v}+v e^{-u} ; u=\ln r, v=s \ln r $$

Step-by-Step Solution

Answer

\(\frac{\partial z}{\partial r} = r^{s-1} + s \ln r \cdot r^{s-1} - \frac{s \ln r}{r^{2}} + \frac{s}{r^{2}}\), \(\frac{\partial z}{\partial s} = (\ln r)^{2} r^{s} + \frac{\ln r}{r}\).

Step 1: Understand the Problem

We need to find the partial derivatives of function \(z\), which is given as \(z = ue^{v} + ve^{-u}\), with respect to \(r\) and \(s\). We are also given the substitutions \(u = \ln r\) and \(v = s \ln r\).

Step 2: Express z in Terms of r and s

First, substitute \(u = \ln r\) and \(v = s \ln r\) into the expression for \(z\). This gives: \[ z = (\ln r) e^{s \ln r} + (s \ln r) e^{- \ln r}. \]

Step 3: Simplify z's Expression

Rewrite the expression using properties of exponents:- \(e^{s \ln r} = r^{s}\), which comes from \(e^{\ln a} = a\).- \(e^{- \ln r} = \frac{1}{r}\).Thus, \(z = (\ln r) r^{s} + \frac{s \ln r}{r}\).

Step 4: Differentiate z with respect to r

To find \(\partial z / \partial r\), take the partial derivative of \(z\) with respect to \(r\):1. Differentiate \((\ln r)r^{s}\) using the product rule: \[ \frac{d}{dr}((\ln r)r^{s}) = r^{s} \cdot \frac{1}{r} + (\ln r) s r^{s-1}.\]2. Differentiate \(\frac{s \ln r}{r}\) using the quotient rule: \[ -\frac{s}{r^{2}}\cdot \ln r + \frac{s}{r}\cdot \frac{1}{r} = -\frac{s \ln r}{r^{2}} + \frac{s}{r^{2}}.\]Combine the results:\[ \frac{\partial z}{\partial r} = \frac{r^{s}}{r} + (\ln r) s r^{s-1} - \frac{s \ln r}{r^{2}} + \frac{s}{r^{2}}. \]

Step 5: Differentiate z with respect to s

To find \(\partial z / \partial s\), take the partial derivative of \(z\) with respect to \(s\):1. Differentiate \((\ln r)r^{s}\): \[ (\ln r) r^{s} \ln r = (\ln r)^{2} r^{s}.\]2. Differentiate \(\frac{s \ln r}{r}\): \[ \ln r \cdot \frac{1}{r}. \]Add the results:\[ \frac{\partial z}{\partial s} = (\ln r)^{2} r^{s} + \frac{\ln r}{r}. \]

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus that allows us to differentiate composite functions. When dealing with functions that are interdependent, like in our problem where both \( u \) and \( v \) depend on \( r \), the chain rule becomes crucial.Think of it this way: if you have a function \( z = f(g(x)) \), the chain rule states that the derivative of \( z \) with respect to \( x \) is the derivative of \( f \) with respect to \( g(x) \), multiplied by the derivative of \( g(x) \) with respect to \( x \). In formula terms, this is \( \frac{dz}{dx} = f'(g(x)) \cdot g'(x) \).In our scenario:- \( u = \ln r \) and \( v = s \ln r \) determine how \( r \) influences \( z \).- Calculating \( \frac{\partial z}{\partial r} \) involves considering how changes in \( r \) affect \( u \) and \( v \) first, then their combined effect on \( z \).Therefore, use the chain rule when you have nested or composite functions, to bring clarity to how inner functions impact the outer function.

Product Rule

The product rule is essential when you need to differentiate functions that are multiplied together. This rule tells us how to find the derivative of a product of two functions, \( u(x) \) and \( v(x) \).If \( z(x) = u(x) \cdot v(x) \), then the derivative of \( z \) with respect to \( x \) is given by: \[ z'(x) = u'(x)v(x) + u(x)v'(x). \]In our exercise, when calculating \( \frac{\partial z}{\partial r} \), the expression \( (\ln r)r^s \) is a product of \( \ln r \) and \( r^s \). Here’s how:- Differentiate \( \ln r \) resulting in \( \frac{1}{r} \).- Differentiate \( r^s \) to obtain \( s r^{s-1} \).- Apply the product rule: \( r^s \frac{1}{r} + (\ln r) s r^{s-1} \).Utilize the product rule whenever facing multiplication of two differentiable functions. It ensures you account for both parts' rates of change appropriately.

Quotient Rule

The quotient rule is quite useful when differentiating a function represented as one function divided by another. If \( z(x) = \frac{u(x)}{v(x)} \), it states:\[ z'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}. \]Applying this to our problem, for the term \( \frac{s \ln r}{r} \) in \( z \), here is the differentiation process:- Set \( u = s \ln r \) and \( v = r \).- Differentiate \( u * = s \cdot \frac{1}{r} \).- Differentiate \( v \) obtaining \( 1 \).- Apply the quotient rule: \( \frac{s (1 - \ln r)}{r^2} = -\frac{s \ln r}{r^2} + \frac{s}{r^2} \).The quotient rule is vital when you have the division of two functions. It guides you in differentiating both the numerator and the denominator, resulting in an accurate derivative of the quotient function.