Answer:
[tex]\displaystyle \boxed{ \frac{dz}{du} = \frac{e^\big{\frac{u(2u + v)}{v}}}{v} \Bigg[ 4u + v \Bigg] }[/tex]
[tex]\displaystyle \boxed{ \frac{dz}{dv} = \frac{ue^\big{\frac{u(2u + v)}{v}}}{v} \Bigg[ 1 - \frac{2u + v}{v} \Bigg] }[/tex]
General Formulas and Concepts:
Calculus
Differentiation
- Derivatives
- Derivative Notation
Derivative Property [Multiplied Constant]:
[tex]\displaystyle (cu)' = cu'[/tex]
Derivative Property [Addition/Subtraction]:
[tex]\displaystyle (u + v)' = u' + v'[/tex]
Derivative Rule [Basic Power Rule]:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Multivariable Calculus
Partial Derivatives
Partial Derivative Rule [Chain Rule]:
[tex]\displaystyle \frac{dw}{dt} = \frac{\partial w}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial w}{\partial y} \frac{\partial y}{\partial t} + \frac{\partial w}{\partial z} \frac{\partial z}{\partial t}[/tex]
Step-by-step explanation:
Step 1: Define
Identify given.
[tex]\displaystyle x = 2u + v[/tex]
[tex]\displaystyle y = \frac{u}{v}[/tex]
[tex]\displaystyle z = e^{xy}[/tex]
Step 2: Find Derivatives
[tex]\displaystyle \frac{dz}{du}[/tex]:
- [Derivative] Rewrite [Partial Derivative Rule - Chain Rule]:
[tex]\displaystyle \frac{dz}{du} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial u}[/tex]
We can find the partial derivatives by differentiating using basic differentiation rules found under "Calculus":
[tex]\displaystyle\begin{aligned}\frac{\partial z}{\partial x} & = ye^{xy} \\\frac{\partial x}{\partial u} & = 2 \\\frac{\partial z}{\partial y} & = xe^{xy} \\\frac{\partial y}{\partial u} & = \frac{1}{v} \\\end{aligned}[/tex]
Substituting in our partial derivatives and our variables x and y, we can obtain a final derivative:
[tex]\displaystyle\begin{aligned}\frac{dz}{du} & = 2ye^{xy} + \frac{xe^{xy}}{v} \\& = \frac{2ue^\big{\frac{u(2u + v)}{v}}}{v} + \frac{(2u + v)e^\big{\frac{u(2u + v)}{v}}}{v} \\& = \boxed{ \frac{e^\big{\frac{u(2u + v)}{v}}}{v} \Bigg[ 4u + v \Bigg] }\end{aligned}[/tex]
∴ we have found the derivative of z with respect to u.
[tex]\displaystyle \frac{dz}{dv}[/tex]:
- [Derivative] Rewrite [Partial Derivative Rule - Chain Rule]:
[tex]\displaystyle \frac{dz}{dv} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial v} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial v}[/tex]
We can find the partial derivatives by using the same method(s):
[tex]\displaystyle\begin{aligned}\frac{\partial z}{\partial x} & = ye^{xy} \\\frac{\partial x}{\partial v} & = 1 \\ \frac{\partial z}{\partial y} & = xe^{xy} \\\frac{\partial x}{\partial v} & = \frac{-u}{v^2} \\ \end{aligned}[/tex]
Substituting in our partial derivatives and our variables x and y, we can obtain a final derivative:
[tex]\displaystyle\begin{aligned}\frac{dz}{dv} & = ye^{xy} + \frac{-uxe^{xy}}{v^2} \\& = \frac{ue^\big{\frac{u(2u + v)}{v}}}{v} + \frac{-u(2u + v)e^\big{\frac{u(2u + v)}{v}}}{v^2} \\& = \boxed{ \frac{ue^\big{\frac{u(2u + v)}{v}}}{v} \Bigg[ 1 - \frac{2u + v}{v} \Bigg] }\end{aligned}[/tex]
∴ we have found the derivative of z with respect to v.
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Learn more about partial derivatives: https://brainly.com/question/6212480
Learn more about multivariable calculus: https://brainly.com/question/12680404
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Topic: Multivariable Calculus
Unit: Partial Derivatives and Applications
