Problem 39 - Advanced Engineering Mathematics

Question

Use the Chain Rule to find the indicated partial derivatives. $$ z=e^{u v^{2}} ; u=x^{3}, v=x-y^{2} ; \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y} $$

Step-by-Step Solution

Answer

\( \frac{\partial z}{\partial x} = e^{uv^2}(3x^2v^2 + 2uv) \) and \( \frac{\partial z}{\partial y} = -4yuv e^{uv^2} \).

Step 1: 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 \( \frac{\partial z}{\partial y} \).

Step 2: 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 \( y \).

Step 3: 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:\[ \frac{\partial z}{\partial x} = v^2 e^{uv^2} \cdot 3x^2 + 2uv e^{uv^2} \cdot 1. \]

Step 4: 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:\[ \frac{\partial z}{\partial x} = e^{uv^2}(3x^2v^2 + 2uv). \]

Step 5: 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:\[ \frac{\partial z}{\partial y} = 2uv e^{uv^2} \cdot (-2y). \]

Step 6: 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 \( \frac{\partial z}{\partial y} \).

Key Concepts

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:

• \( \frac{\partial z}{\partial u} \): 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 \( y \).

Using these components, we apply the Chain Rule to find \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \). It involves multiplying and adding these simple derivatives together.

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:

• 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.

In this context, knowing how to switch between variables and their interactions is critical. It helps to analyze how the output \( z \) behaves with changing \( x \) and \( y \), reflecting the essence of multivariable functions.

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:

• 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.

Remember, differentiation highlights critical aspects like rates of change and tangent lines in multidimensional spaces. It's a fundamental tool for analyzing functions, solving problems, and understanding dynamic systems. The careful step-by-step calculation reveals this and forms the basis for more advanced mathematical concepts.