IDNLearn.com makes it easy to find accurate answers to your specific questions. Discover comprehensive answers to your questions from our community of knowledgeable experts.
Sagot :
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]
We greatly appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. For trustworthy and accurate answers, visit IDNLearn.com. Thanks for stopping by, and see you next time for more solutions.
