Answer :
Sure! Let's solve this step-by-step to find the partial derivatives [tex]\(\frac{\partial f}{\partial u}\)[/tex], [tex]\(\frac{\partial f}{\partial v}\)[/tex], and [tex]\(\frac{\partial f}{\partial w}\)[/tex] for the given function.
### Step 1: Define the function [tex]\(f(x, y, z)\)[/tex]
Given:
[tex]\[ f(x, y, z) = 9x - 7y^2 + 3z^2 \][/tex]
### Step 2: Substitute [tex]\(x\)[/tex], [tex]\(y\)[/tex], and [tex]\(z\)[/tex] with their expressions in terms of [tex]\(u\)[/tex], [tex]\(v\)[/tex], and [tex]\(w\)[/tex]
[tex]\[ x = \sqrt{u + v} \][/tex]
[tex]\[ y = (u + w)\ln(v) \][/tex]
[tex]\[ z = 2 - v + 5w \][/tex]
### Step 3: Substitute these expressions into [tex]\(f\)[/tex]
So, the function [tex]\(f\)[/tex] in terms of [tex]\(u\)[/tex], [tex]\(v\)[/tex], and [tex]\(w\)[/tex] becomes:
[tex]\[ f(u, v, w) = 9\sqrt{u + v} - 7((u + w) \ln(v))^2 + 3(2 - v + 5w)^2 \][/tex]
### Step 4: Calculate the partial derivatives
#### Partial derivative with respect to [tex]\(u\)[/tex]:
[tex]\[ \frac{\partial f}{\partial u} = \frac{\partial }{\partial u} \left( 9\sqrt{u + v} - 7((u + w) \ln(v))^2 + 3(2 - v + 5w)^2 \right) \][/tex]
The only terms in [tex]\(f\)[/tex] that depend on [tex]\(u\)[/tex] are [tex]\(9\sqrt{u + v}\)[/tex] and [tex]\(-7((u + w)\ln(v))^2\)[/tex].
[tex]\[ \frac{\partial }{\partial u} \left( 9\sqrt{u + v} \right) = \frac{9}{2\sqrt{u + v}} \][/tex]
[tex]\[ \frac{\partial }{\partial u} \left( -7((u + w)\ln(v))^2 \right) = -7 \cdot 2((u + w)\ln(v)) \cdot \ln(v) = -14(u + w)\ln(v)^2 \][/tex]
So,
[tex]\[ \frac{\partial f}{\partial u} = \frac{9}{2\sqrt{u + v}} - 14(u + w)\ln(v)^2 \][/tex]
#### Partial derivative with respect to [tex]\(v\)[/tex]:
[tex]\[ \frac{\partial f}{\partial v} = \frac{\partial }{\partial v} \left( 9\sqrt{u + v} - 7((u + w) \ln(v))^2 + 3(2 - v + 5w)^2 \right) \][/tex]
We have contributions from all three terms:
[tex]\[ \frac{\partial }{\partial v} \left( 9\sqrt{u + v} \right) = \frac{9}{2\sqrt{u + v}} \][/tex]
[tex]\[ \frac{\partial }{\partial v} \left( -7((u + w)\ln(v))^2 \right) = -7 \left( 2(u + w)\ln(v) \cdot \frac{u + w}{v} \right) = -14(u + w)^2\frac{\ln(v)}{v} \][/tex]
[tex]\[ \frac{\partial }{\partial v} \left( 3(2 - v + 5w)^2 \right) = 3 \cdot 2(2 - v + 5w) \cdot (-1) = -6(2 - v + 5w) = -12 + 6v - 30w \][/tex]
Combining these:
[tex]\[ \frac{\partial f}{\partial v} = \frac{9}{2\sqrt{u + v}} - 14(u + w)^2\frac{\ln(v)}{v} + 6v - 30w - 12 \][/tex]
#### Partial derivative with respect to [tex]\(w\)[/tex]:
[tex]\[ \frac{\partial f}{\partial w} = \frac{\partial }{\partial w} \left( 9\sqrt{u + v} - 7((u + w) \ln(v))^2 + 3(2 - v + 5w)^2 \right) \][/tex]
Again, consider each term:
[tex]\[ \frac{\partial }{\partial w} \left( 9\sqrt{u + v} \right) = 0 \][/tex]
[tex]\[ \frac{\partial }{\partial w} \left( -7((u + w)\ln(v))^2 \right) = -7 \cdot 2(u + w)\ln(v)^2 = -14(u + w)\ln(v)^2 \][/tex]
[tex]\[ \frac{\partial }{\partial w} \left( 3(2 - v + 5w)^2 \right) = 3 \cdot 2(2 - v + 5w) \cdot 5 = 30(2 - v + 5w) = 60 - 30v + 150w \][/tex]
So,
[tex]\[ \frac{\partial f}{\partial w} = -14(u + w)\ln(v)^2 + 60 - 30v + 150w \][/tex]
### Final Result:
Combining all, we have the partial derivatives:
[tex]\[ \frac{\partial f}{\partial u} = \frac{9}{2\sqrt{u + v}} - 14(u + w)\ln(v)^2 \][/tex]
[tex]\[ \frac{\partial f}{\partial v} = \frac{9}{2\sqrt{u + v}} - 14(u + w)^2\frac{\ln(v)}{v} + 6v - 30w - 12 \][/tex]
[tex]\[ \frac{\partial f}{\partial w} = -14(u + w)\ln(v)^2 + 60 - 30v + 150w \][/tex]
### Step 1: Define the function [tex]\(f(x, y, z)\)[/tex]
Given:
[tex]\[ f(x, y, z) = 9x - 7y^2 + 3z^2 \][/tex]
### Step 2: Substitute [tex]\(x\)[/tex], [tex]\(y\)[/tex], and [tex]\(z\)[/tex] with their expressions in terms of [tex]\(u\)[/tex], [tex]\(v\)[/tex], and [tex]\(w\)[/tex]
[tex]\[ x = \sqrt{u + v} \][/tex]
[tex]\[ y = (u + w)\ln(v) \][/tex]
[tex]\[ z = 2 - v + 5w \][/tex]
### Step 3: Substitute these expressions into [tex]\(f\)[/tex]
So, the function [tex]\(f\)[/tex] in terms of [tex]\(u\)[/tex], [tex]\(v\)[/tex], and [tex]\(w\)[/tex] becomes:
[tex]\[ f(u, v, w) = 9\sqrt{u + v} - 7((u + w) \ln(v))^2 + 3(2 - v + 5w)^2 \][/tex]
### Step 4: Calculate the partial derivatives
#### Partial derivative with respect to [tex]\(u\)[/tex]:
[tex]\[ \frac{\partial f}{\partial u} = \frac{\partial }{\partial u} \left( 9\sqrt{u + v} - 7((u + w) \ln(v))^2 + 3(2 - v + 5w)^2 \right) \][/tex]
The only terms in [tex]\(f\)[/tex] that depend on [tex]\(u\)[/tex] are [tex]\(9\sqrt{u + v}\)[/tex] and [tex]\(-7((u + w)\ln(v))^2\)[/tex].
[tex]\[ \frac{\partial }{\partial u} \left( 9\sqrt{u + v} \right) = \frac{9}{2\sqrt{u + v}} \][/tex]
[tex]\[ \frac{\partial }{\partial u} \left( -7((u + w)\ln(v))^2 \right) = -7 \cdot 2((u + w)\ln(v)) \cdot \ln(v) = -14(u + w)\ln(v)^2 \][/tex]
So,
[tex]\[ \frac{\partial f}{\partial u} = \frac{9}{2\sqrt{u + v}} - 14(u + w)\ln(v)^2 \][/tex]
#### Partial derivative with respect to [tex]\(v\)[/tex]:
[tex]\[ \frac{\partial f}{\partial v} = \frac{\partial }{\partial v} \left( 9\sqrt{u + v} - 7((u + w) \ln(v))^2 + 3(2 - v + 5w)^2 \right) \][/tex]
We have contributions from all three terms:
[tex]\[ \frac{\partial }{\partial v} \left( 9\sqrt{u + v} \right) = \frac{9}{2\sqrt{u + v}} \][/tex]
[tex]\[ \frac{\partial }{\partial v} \left( -7((u + w)\ln(v))^2 \right) = -7 \left( 2(u + w)\ln(v) \cdot \frac{u + w}{v} \right) = -14(u + w)^2\frac{\ln(v)}{v} \][/tex]
[tex]\[ \frac{\partial }{\partial v} \left( 3(2 - v + 5w)^2 \right) = 3 \cdot 2(2 - v + 5w) \cdot (-1) = -6(2 - v + 5w) = -12 + 6v - 30w \][/tex]
Combining these:
[tex]\[ \frac{\partial f}{\partial v} = \frac{9}{2\sqrt{u + v}} - 14(u + w)^2\frac{\ln(v)}{v} + 6v - 30w - 12 \][/tex]
#### Partial derivative with respect to [tex]\(w\)[/tex]:
[tex]\[ \frac{\partial f}{\partial w} = \frac{\partial }{\partial w} \left( 9\sqrt{u + v} - 7((u + w) \ln(v))^2 + 3(2 - v + 5w)^2 \right) \][/tex]
Again, consider each term:
[tex]\[ \frac{\partial }{\partial w} \left( 9\sqrt{u + v} \right) = 0 \][/tex]
[tex]\[ \frac{\partial }{\partial w} \left( -7((u + w)\ln(v))^2 \right) = -7 \cdot 2(u + w)\ln(v)^2 = -14(u + w)\ln(v)^2 \][/tex]
[tex]\[ \frac{\partial }{\partial w} \left( 3(2 - v + 5w)^2 \right) = 3 \cdot 2(2 - v + 5w) \cdot 5 = 30(2 - v + 5w) = 60 - 30v + 150w \][/tex]
So,
[tex]\[ \frac{\partial f}{\partial w} = -14(u + w)\ln(v)^2 + 60 - 30v + 150w \][/tex]
### Final Result:
Combining all, we have the partial derivatives:
[tex]\[ \frac{\partial f}{\partial u} = \frac{9}{2\sqrt{u + v}} - 14(u + w)\ln(v)^2 \][/tex]
[tex]\[ \frac{\partial f}{\partial v} = \frac{9}{2\sqrt{u + v}} - 14(u + w)^2\frac{\ln(v)}{v} + 6v - 30w - 12 \][/tex]
[tex]\[ \frac{\partial f}{\partial w} = -14(u + w)\ln(v)^2 + 60 - 30v + 150w \][/tex]
