Chapter 9: Problem 39
Use the Chain Rule to find the indicated partial derivatives. $$ z=e^{u v^{2}} ; u=x^{3}, v=x-y^{2} ; \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y} $$
Short Answer
Expert verified
\( \frac{\partial z}{\partial x} = e^{uv^2}(3x^2v^2 + 2uv) \) and \( \frac{\partial z}{\partial y} = -4yuv e^{uv^2} \).
Step by step solution
01
Identify the Function Relationships
We have the function given as \( z = e^{uv^2} \). The variables \( u \) and \( v \) are defined in terms of \( x \) and \( y \) as \( u = x^3 \) and \( v = x - y^2 \). Our goal is to find \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \).
02
Expression for \( z \) in terms of \( x \) and \( y \)
Substitute \( u = x^3 \) and \( v = x - y^2 \) into the expression for \( z \): \[ z = e^{(x^3)(x-y^2)^2}. \] Now, \( z \) is expressed in terms of \( x \) and \( y \).
03
Apply the Chain Rule for \( \frac{\partial z}{\partial x} \)
By the Chain Rule, \( \frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial x} \). First, find these partial derivatives:1. \( \frac{\partial z}{\partial u} = v^2 e^{uv^2} \) since \( z = e^{uv^2} \).2. \( \frac{\partial z}{\partial v} = 2uv e^{uv^2} \).3. \( \frac{\partial u}{\partial x} = 3x^2 \).4. \( \frac{\partial v}{\partial x} = 1 \).Now substitute them into the equation:\[ \frac{\partial z}{\partial x} = v^2 e^{uv^2} \cdot 3x^2 + 2uv e^{uv^2} \cdot 1. \]
04
Simplification of \( \frac{\partial z}{\partial x} \)
Simplify the expression:\[ \frac{\partial z}{\partial x} = 3x^2v^2 e^{uv^2} + 2uv e^{uv^2}. \]Factoring out \( e^{uv^2} \), we get:\[ \frac{\partial z}{\partial x} = e^{uv^2}(3x^2v^2 + 2uv). \]
05
Apply the Chain Rule for \( \frac{\partial z}{\partial y} \)
By the Chain Rule, \( \frac{\partial z}{\partial y} = \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial y} \). First, find the remaining partial derivatives:1. From Step 3, \( \frac{\partial z}{\partial v} = 2uv e^{uv^2} \).2. \( \frac{\partial v}{\partial y} = -2y \).Now substitute them into the equation:\[ \frac{\partial z}{\partial y} = 2uv e^{uv^2} \cdot (-2y). \]
06
Simplification of \( \frac{\partial z}{\partial y} \)
Simplify the expression:\[ \frac{\partial z}{\partial y} = -4yuv e^{uv^2}. \] This is the final form of \( \frac{\partial z}{\partial y} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chain Rule
The Chain Rule is a powerful tool in calculus, especially when dealing with functions of multiple variables. It allows us to differentiate composite functions. When you have a function composed of two or more functions, the Chain Rule helps find the derivative efficiently.
In the exercise, we have a function \( z = e^{uv^2} \) where \( u = x^3 \) and \( v = x - y^2 \). To find the partial derivatives of \( z \) with respect to \( x \) and \( y \), we use the Chain Rule. It breaks down the differentiation of complicated functions by expressing derivatives in terms of simpler ones.
Components of the Chain Rule:
In the exercise, we have a function \( z = e^{uv^2} \) where \( u = x^3 \) and \( v = x - y^2 \). To find the partial derivatives of \( z \) with respect to \( x \) and \( y \), we use the Chain Rule. It breaks down the differentiation of complicated functions by expressing derivatives in terms of simpler ones.
Components of the Chain Rule:
- \( \frac{\partial z}{\partial u} \): Derivative of \( z \) with respect to \( u \).
- \( \frac{\partial z}{\partial v} \): Derivative of \( z \) with respect to \( v \).
- \( \frac{\partial u}{\partial x} \) and \( \frac{\partial v}{\partial x} \): Derivative of \( u \) and \( v \) with respect to \( x \).
- \( \frac{\partial v}{\partial y} \): Derivative of \( v \) with respect to \( y \).
Multivariable Calculus
Multivariable Calculus extends the concepts of calculus to functions of more than one variable. Unlike single-variable calculus, where you only differentiate with respect to one variable, multivariable calculus involves partial derivatives.
Partial derivatives measure the change in a function with respect to one variable while keeping the other variables constant. In our exercise, \( z = e^{uv^2} \) depends on both \( x \) and \( y \) through \( u \) and \( v \), which is typical in multivariable calculus.
Key Points in Multivariable Calculus:
Partial derivatives measure the change in a function with respect to one variable while keeping the other variables constant. In our exercise, \( z = e^{uv^2} \) depends on both \( x \) and \( y \) through \( u \) and \( v \), which is typical in multivariable calculus.
Key Points in Multivariable Calculus:
- Understanding how each variable influences the function.
- Applying techniques like the Chain Rule to break down complex derivatives.
- Visualizing each derivative as describing a slope in a particular direction.
Differentiation
Differentiation is the process of finding a derivative of a function, which reveals how it changes and behaves. For multivariable functions, as in our example \( z = e^{uv^2} \), we perform partial differentiation to understand these changes relative to each variable.
In steps, this involves:
In steps, this involves:
- Identifying the relationship between variables, as \( z \) in terms of \( u \) and \( v \).
- Calculating partial derivatives \( \frac{\partial z}{\partial u} \) and \( \frac{\partial z}{\partial v} \).
- Using these derivatives to find \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \) through the Chain Rule.