Problem 22 - Advanced Engineering Mathematics

Question

Find the first partial derivatives of the given function. $$ z=e^{x^{2} \tan ^{-1} y^{2}} $$

Step-by-Step Solution

Answer

The partial derivatives are \(\frac{\partial z}{\partial x} = e^{x^2 \tan^{-1}(y^2)} \cdot 2x \tan^{-1}(y^2)\) and \(\frac{\partial z}{\partial y} = e^{x^2 \tan^{-1}(y^2)} \cdot \frac{2x^2 y}{1+y^4}\).

Step 1: Identify Variables and Function

The given function is \( z = e^{x^2 \tan^{-1}(y^2)} \). Here, \(x\) and \(y\) are the independent variables, and \(z\) is the dependent variable. We need to find the partial derivatives of \(z\) with respect to \(x\) and \(y\).

Step 2: Differentiate with Respect to x

To find the partial derivative of \(z\) with respect to \(x\), we use the chain rule. The function can be rewritten as \(z = e^u\) where \(u = x^2 \tan^{-1}(y^2)\). Thus, \(\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x}\).Calculate \(\frac{\partial z}{\partial u} = e^u\) since the derivative of \(e^u\) with respect to \(u\) is \(e^u\).Next, find \(\frac{\partial u}{\partial x} = \frac{\partial}{\partial x}(x^2 \tan^{-1}(y^2)) = 2x \tan^{-1}(y^2)\) (applying product rule where \(\tan^{-1}(y^2)\) is constant).The partial derivative \(\frac{\partial z}{\partial x} = e^{x^2 \tan^{-1}(y^2)} \cdot 2x \tan^{-1}(y^2)\).

Step 3: Differentiate with Respect to y

Now, we need to find the partial derivative of \(z\) with respect to \(y\) using the chain rule again.We have \(\frac{\partial z}{\partial y} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial y}\).We already know \(\frac{\partial z}{\partial u} = e^u\).For \(\frac{\partial u}{\partial y} = \frac{\partial}{\partial y}(x^2 \tan^{-1}(y^2)) = x^2 \cdot \frac{d}{dy}(\tan^{-1}(y^2)) = x^2 \cdot \frac{2y}{1+(y^2)^2}\), where the derivative of \(\tan^{-1}(v)\) is \(\frac{1}{1+v^2}\) and \(v = y^2\).Thus, \(\frac{\partial z}{\partial y} = e^{x^2 \tan^{-1}(y^2)} \cdot x^2 \cdot \frac{2y}{1+y^4}\).

Key Concepts

The Chain Rule in Partial Derivatives

When dealing with partial derivatives, the chain rule is an essential tool. It helps us to differentiate functions that are composed of other functions. The idea is to express a derivative in terms of intermediate variables. Think of it as a tool to "break" a function down into simpler parts.
In a multivariable setting, the chain rule allows us to find how a function changes with respect to one of its variables while accounting for how this depends on one or more other variables.

• For a function \(z = f(x, y)\), if we express \(z\) in terms of another variable \(u\), such as \(z = e^u\) where \(u = x^2 \tan^{-1}(y^2)\), we can find partial derivatives \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\).
• If \(z = g(u)\) and \(u = h(x, y)\), the chain rule says \(\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x}\) and similarly for \(y\).

This systematic approach simplifies the calculation of derivatives of composite functions. It is like taking apart a puzzle, piece by piece.

Applying the Product Rule

The product rule comes into play when we need to differentiate a product of two functions. It states that the derivative of a product of two functions is the first function times the derivative of the second function plus the second function times the derivative of the first.
In the context of partial derivatives, the product rule helps simplify the differentiation process when you have terms that are multiplied together. Let's look at an example for clarity:

• Consider \(u = x^2 \tan^{-1}(y^2)\). Here, \(x^2\) and \(\tan^{-1}(y^2)\) are multiplied, so we use the product rule.
• According to the product rule, \(\frac{\partial u}{\partial x}\) is obtained by differentiating \(x^2\) with respect to \(x\) (which results in \(2x\)) and treating \(\tan^{-1}(y^2)\) as a constant, then reversing the roles.

It is crucial to identify when and where the product rule applies, especially in functions where components are intertwined by multiplication.

Understanding Partial Differentiation

Partial differentiation allows us to find the rate of change of a function with respect to one variable at a time, while keeping other variables constant. It is as if we are cutting through the layers of the function to focus on one specific variable.
This approach is particularly useful when working with functions of multiple variables, as it tells us how sensitive the function is to changes in each variable individually. Let's explore how you use partial differentiation:

• The partial derivative of \(z = f(x, y)\) with respect to \(x\) is denoted \(\frac{\partial z}{\partial x}\), which indicates how \(z\) changes with small changes in \(x\), holding \(y\) constant.
• Similarly, \(\frac{\partial z}{\partial y}\) considers \(x\) constant.

This method is central to fields like physics and economics, where multivariable functions often arise, and understanding the influence of each individual factor is vital. Mastering partial differentiation opens up a deeper understanding of how complex systems behave.