Answer :
Certainly! Let's go through the process step by step to find the partial derivatives [tex]\(\frac{\partial z}{\partial u}\)[/tex] and [tex]\(\frac{\partial z}{\partial v}\)[/tex].
First, let's write down the given expressions and the objective:
Given:
[tex]\[ z = \frac{1}{4y} \ln(x) \][/tex]
[tex]\[ x = \sqrt{uv} \][/tex]
[tex]\[ y = \frac{6v}{u} \][/tex]
We want to find:
[tex]\[ \frac{\partial z}{\partial u} \][/tex]
[tex]\[ \frac{\partial z}{\partial v} \][/tex]
---
### Step 1: Express [tex]\(z\)[/tex] in terms of [tex]\(u\)[/tex] and [tex]\(v\)[/tex]
Substitute [tex]\(x = \sqrt{uv}\)[/tex] and [tex]\(y = \frac{6v}{u}\)[/tex] into the expression for [tex]\(z\)[/tex]:
[tex]\[ z = \frac{1}{4 \left(\frac{6v}{u}\right)} \ln\left(\sqrt{uv}\right) \][/tex]
### Step 2: Simplify the expression
First, simplify the denominator [tex]\( 4 \left(\frac{6v}{u}\right) \)[/tex]:
[tex]\[ 4 \left(\frac{6v}{u}\right) = \frac{24v}{u} \][/tex]
Next, simplify the natural logarithm expression [tex]\(\ln(\sqrt{uv})\)[/tex]:
[tex]\[ \ln(\sqrt{uv}) = \ln((uv)^{1/2}) = \frac{1}{2} \ln(uv) = \frac{1}{2} (\ln u + \ln v) \][/tex]
Thus,
[tex]\[ z = \frac{1}{\frac{24v}{u}} \cdot \frac{1}{2} (\ln u + \ln v) = \frac{u}{24v} \cdot \frac{1}{2} (\ln u + \ln v) = \frac{u}{48v} (\ln u + \ln v) \][/tex]
Rewriting:
[tex]\[ z = \frac{u \ln u}{48v} + \frac{u \ln v}{48v} \][/tex]
### Step 3: Find [tex]\(\frac{\partial z}{\partial u}\)[/tex]
To find [tex]\(\frac{\partial z}{\partial u}\)[/tex]:
[tex]\[ z = \frac{u \ln u}{48v} + \frac{u \ln v}{48v} \][/tex]
First, consider the term [tex]\(\frac{u \ln u}{48v}\)[/tex]:
[tex]\[ \frac{\partial}{\partial u} \left(\frac{u \ln u}{48v}\right) = \frac{1}{48v} \left(\ln u + 1\right) \][/tex]
Next, consider the term [tex]\(\frac{u \ln v}{48v}\)[/tex]:
[tex]\[ \frac{\partial}{\partial u} \left(\frac{u \ln v}{48v}\right) = \frac{\ln v}{48v} \][/tex]
Thus,
[tex]\[ \frac{\partial z}{\partial u} = \frac{1}{48v} \left(\ln u + 1\right) + \frac{\ln v}{48v} = \frac{1}{48v} \left(\ln u + \ln v + 1\right) \][/tex]
Given [tex]\( \ln u + \ln v = \ln(uv) \)[/tex]:
[tex]\[ \frac{\partial z}{\partial u} = \frac{1 + \ln(uv)}{48v} \][/tex]
---
### Step 4: Find [tex]\(\frac{\partial z}{\partial v}\)[/tex]
To find [tex]\(\frac{\partial z}{\partial v}\)[/tex]:
[tex]\[ z = \frac{u \ln u}{48v} + \frac{u \ln v}{48v} \][/tex]
First, consider the term [tex]\(\frac{u \ln u}{48v}\)[/tex]:
[tex]\[ \frac{\partial}{\partial v} \left(\frac{u \ln u}{48v}\right) = \frac{u \ln u}{48} \cdot \left(-\frac{1}{v^2}\right) = -\frac{u \ln u}{48 v^2} \][/tex]
Next, consider the term [tex]\(\frac{u \ln v}{48v}\)[/tex]:
[tex]\[ \frac{\partial}{\partial v} \left(\frac{u \ln v}{48v}\right) = \frac{u}{48} \cdot \frac{1}{v} - \frac{u \ln v}{48} \cdot \frac{1}{v^2} = \frac{u}{48v^2} - \frac{u \ln v}{48 v^2} \][/tex]
Thus,
[tex]\[ \frac{\partial z}{\partial v} = -\frac{u \ln u}{48 v^2} + \frac{u}{48v^2} - \frac{u \ln v}{48 v^2} = \frac{u (1 - \ln u - \ln v)}{48 v^2} \][/tex]
Given [tex]\( \ln u + \ln v = \ln(uv) \)[/tex]:
[tex]\[ \frac{\partial z}{\partial v} = \frac{u (1 - \ln(uv))}{48 v^2} \][/tex]
In conclusion, the partial derivatives are:
[tex]\[ \frac{\partial z}{\partial u} = \frac{1 + \ln(uv)}{48v} \][/tex]
[tex]\[ \frac{\partial z}{\partial v} = \frac{u (1 - \ln(uv))}{48 v^2} \][/tex]
First, let's write down the given expressions and the objective:
Given:
[tex]\[ z = \frac{1}{4y} \ln(x) \][/tex]
[tex]\[ x = \sqrt{uv} \][/tex]
[tex]\[ y = \frac{6v}{u} \][/tex]
We want to find:
[tex]\[ \frac{\partial z}{\partial u} \][/tex]
[tex]\[ \frac{\partial z}{\partial v} \][/tex]
---
### Step 1: Express [tex]\(z\)[/tex] in terms of [tex]\(u\)[/tex] and [tex]\(v\)[/tex]
Substitute [tex]\(x = \sqrt{uv}\)[/tex] and [tex]\(y = \frac{6v}{u}\)[/tex] into the expression for [tex]\(z\)[/tex]:
[tex]\[ z = \frac{1}{4 \left(\frac{6v}{u}\right)} \ln\left(\sqrt{uv}\right) \][/tex]
### Step 2: Simplify the expression
First, simplify the denominator [tex]\( 4 \left(\frac{6v}{u}\right) \)[/tex]:
[tex]\[ 4 \left(\frac{6v}{u}\right) = \frac{24v}{u} \][/tex]
Next, simplify the natural logarithm expression [tex]\(\ln(\sqrt{uv})\)[/tex]:
[tex]\[ \ln(\sqrt{uv}) = \ln((uv)^{1/2}) = \frac{1}{2} \ln(uv) = \frac{1}{2} (\ln u + \ln v) \][/tex]
Thus,
[tex]\[ z = \frac{1}{\frac{24v}{u}} \cdot \frac{1}{2} (\ln u + \ln v) = \frac{u}{24v} \cdot \frac{1}{2} (\ln u + \ln v) = \frac{u}{48v} (\ln u + \ln v) \][/tex]
Rewriting:
[tex]\[ z = \frac{u \ln u}{48v} + \frac{u \ln v}{48v} \][/tex]
### Step 3: Find [tex]\(\frac{\partial z}{\partial u}\)[/tex]
To find [tex]\(\frac{\partial z}{\partial u}\)[/tex]:
[tex]\[ z = \frac{u \ln u}{48v} + \frac{u \ln v}{48v} \][/tex]
First, consider the term [tex]\(\frac{u \ln u}{48v}\)[/tex]:
[tex]\[ \frac{\partial}{\partial u} \left(\frac{u \ln u}{48v}\right) = \frac{1}{48v} \left(\ln u + 1\right) \][/tex]
Next, consider the term [tex]\(\frac{u \ln v}{48v}\)[/tex]:
[tex]\[ \frac{\partial}{\partial u} \left(\frac{u \ln v}{48v}\right) = \frac{\ln v}{48v} \][/tex]
Thus,
[tex]\[ \frac{\partial z}{\partial u} = \frac{1}{48v} \left(\ln u + 1\right) + \frac{\ln v}{48v} = \frac{1}{48v} \left(\ln u + \ln v + 1\right) \][/tex]
Given [tex]\( \ln u + \ln v = \ln(uv) \)[/tex]:
[tex]\[ \frac{\partial z}{\partial u} = \frac{1 + \ln(uv)}{48v} \][/tex]
---
### Step 4: Find [tex]\(\frac{\partial z}{\partial v}\)[/tex]
To find [tex]\(\frac{\partial z}{\partial v}\)[/tex]:
[tex]\[ z = \frac{u \ln u}{48v} + \frac{u \ln v}{48v} \][/tex]
First, consider the term [tex]\(\frac{u \ln u}{48v}\)[/tex]:
[tex]\[ \frac{\partial}{\partial v} \left(\frac{u \ln u}{48v}\right) = \frac{u \ln u}{48} \cdot \left(-\frac{1}{v^2}\right) = -\frac{u \ln u}{48 v^2} \][/tex]
Next, consider the term [tex]\(\frac{u \ln v}{48v}\)[/tex]:
[tex]\[ \frac{\partial}{\partial v} \left(\frac{u \ln v}{48v}\right) = \frac{u}{48} \cdot \frac{1}{v} - \frac{u \ln v}{48} \cdot \frac{1}{v^2} = \frac{u}{48v^2} - \frac{u \ln v}{48 v^2} \][/tex]
Thus,
[tex]\[ \frac{\partial z}{\partial v} = -\frac{u \ln u}{48 v^2} + \frac{u}{48v^2} - \frac{u \ln v}{48 v^2} = \frac{u (1 - \ln u - \ln v)}{48 v^2} \][/tex]
Given [tex]\( \ln u + \ln v = \ln(uv) \)[/tex]:
[tex]\[ \frac{\partial z}{\partial v} = \frac{u (1 - \ln(uv))}{48 v^2} \][/tex]
In conclusion, the partial derivatives are:
[tex]\[ \frac{\partial z}{\partial u} = \frac{1 + \ln(uv)}{48v} \][/tex]
[tex]\[ \frac{\partial z}{\partial v} = \frac{u (1 - \ln(uv))}{48 v^2} \][/tex]
