Question

Find all first partial derivatives of each function. \(f(x, y)=y \cos \left(x^{2}+y^{2}\right)\)

Short answer

\( \frac{\partial f}{\partial x} = -2xy \sin(x^2 + y^2) \) and \( \frac{\partial f}{\partial y} = \cos(x^2 + y^2) - 2y^2 \sin(x^2 + y^2) \).

Step-by-step solution

Identify the Function Components

We have the function \( f(x, y) = y \cos(x^2 + y^2) \). Here, \( y \) represents a multiplier to the cosine function, and \( x^2 + y^2 \) is an expression inside the cosine that depends on both variables \( x \) and \( y \). Our task is to find the partial derivatives with respect to \( x \) and \( y \).

Partial Derivative with respect to \( x \)

To find \( \frac{\partial f}{\partial x} \), treat \( y \) as a constant. Use the chain rule: the derivative of \( \cos(u) \) with respect to \( u \) is \( -\sin(u) \); then, multiply by the derivative of \( u=x^2+y^2 \) with respect to \( x \), which is \( 2x \). Thus, \( \frac{\partial f}{\partial x} = y \cdot (-\sin(x^2 + y^2)) \cdot 2x = -2xy \sin(x^2 + y^2) \).

Partial Derivative with respect to \( y \)

To find \( \frac{\partial f}{\partial y} \), apply the product rule. Here, differentiate \( y \cdot \cos(x^2 + y^2) \) as a product of two functions:\( y \) and \( \cos(x^2 + y^2) \). First, differentiate the first part \( y \): \( \frac{d}{dy}(y) = 1 \), giving \( \cos(x^2 + y^2) \). Then, use the chain rule on \( \cos(x^2 + y^2) \), which gives \( -\sin(x^2 + y^2) \cdot 2y \). Combined, we have \( \frac{\partial f}{\partial y} = \cos(x^2 + y^2) - 2y^2 \sin(x^2 + y^2) \).

Key concepts

Chain Rule

The chain rule is an essential concept in calculus, especially when dealing with functions nested inside other functions. It allows us to find derivatives of composite functions.
For example, consider the function \(f(x, y) = y \cos(x^2 + y^2)\). To find the partial derivative with respect to \(x\), we treat \(y\) as a constant. The expression inside the cosine, \(x^2 + y^2\), requires us to use the chain rule.
This is done by first taking the derivative of the outer function, \(\cos(u)\), with respect to \(u\), yielding \(-\sin(u)\). Then, multiply by the derivative of the inner function (\(u = x^2 + y^2\)) with respect to \(x\), which is \(2x\). This combination gives:

• \(-\sin(x^2 + y^2)\)
• \(2x\)
  

Multiplying these, and including the constant \(y\), the result is \(-2xy \sin(x^2 + y^2)\). The chain rule efficiently handles the derivative of such composite functions.

Product Rule

The product rule is instrumental when you have a product of two functions and you need to find their derivative. It helps in breaking down the task into manageable parts.
Consider again the function \(f(x, y) = y \cos(x^2 + y^2)\). For the partial derivative with respect to \(y\), we use the product rule.
The function is a product of \(y\) and \(\cos(x^2 + y^2)\). The product rule states that to find the derivative of a product \(u \times v\), apply:

• \(\frac{du}{dy} v + u \frac{dv}{dy}\)

For \(\frac{d}{dy}(y) = 1\), this gives us \( \cos(x^2 + y^2)\). For the second part, use the chain rule on \(\cos(x^2 + y^2)\), resulting in \(-2y \sin(x^2 + y^2)\).
Combining these gives the derivative \(\cos(x^2 + y^2) - 2y^2 \sin(x^2 + y^2)\). The product rule provides clarity in differentiating functions that are products of two or more functions.

Trigonometric Functions

Trigonometric functions, such as \(\sin\) and \(\cos\), frequently appear in calculus exercises. They are periodic functions that model repetitive behavior.
In the context of calculus, one of their key characteristics is their derivatives:

• The derivative of \(\sin(u)\) is \(\cos(u)\)
• The derivative of \(\cos(u)\) is \(-\sin(u)\)
  

These rules are crucial when tackling problems that involve trigonometric components. In our exercise, both \(\sin\) and \(\cos\) play integral roles in finding partial derivatives. The function \(f(x, y) = y \cos(x^2 + y^2)\) utilizes \(\cos(x^2 + y^2)\), necessitating an understanding of its derivative as part of the chain rule and product rule applications.
Knowing how to differentiate trigonometric functions helps solve complex calculus problems efficiently.