Problem 26 - Calculus Early Transcendentals

Question

\(25-30\) Find the differential of the function. $$ v=y \cos x y $$

Step-by-Step Solution

Answer

The differential is \( dv = (-y^2 \sin(xy))dx + (\cos(xy) - xy \sin(xy))dy \).

Step 1: Identify the Variables

The given function is \( v = y \cos(xy) \). This is a function of two variables: \( x \) and \( y \). To find the differential, we need to compute the partial derivatives with respect to both \( x \) and \( y \).

Step 2: Calculate \( \frac{\partial v}{\partial x} \)

Using the product rule, first take the derivative of \( y \cos(xy) \) with respect to \( x \). Here, \( y \) is treated as a constant.Apply the chain rule as follows:\[ \frac{\partial v}{\partial x} = y \cdot \frac{d}{dx}[\cos(xy)] = y \cdot (-\sin(xy)) \cdot \left( y \right) = -y^2 \sin(xy) \]

Step 3: Calculate \( \frac{\partial v}{\partial y} \)

Similarly, apply the product rule with respect to \( y \), treating \( x \) as a constant.Part 1: \( \frac{d}{dy}[y] = 1 \), so that part contributes:\[ 1 \cdot \cos(xy) = \cos(xy) \]Part 2: Again use the chain rule:\[ y \cdot \frac{d}{dy}[\cos(xy)] = y \cdot (-\sin(xy)) \cdot (x) = -xy \sin(xy) \]Thus, combining both parts, we have:\[ \frac{\partial v}{\partial y} = \cos(xy) - xy \sin(xy) \]

Step 4: Write the Differential dv

The differential \( dv \) is given by the formula:\[ dv = \frac{\partial v}{\partial x}dx + \frac{\partial v}{\partial y}dy \]Substituting the derivatives from Steps 2 and 3:\[ dv = (-y^2 \sin(xy))dx + (\cos(xy) - xy \sin(xy))dy \]

Key Concepts

Partial Derivatives

Partial derivatives help us understand how a function changes with respect to one of its variables while keeping the others constant. In our given function, \( v = y \cos(xy) \), the partial derivatives \( \frac{\partial v}{\partial x} \) and \( \frac{\partial v}{\partial y} \) show us how the function \( v \) changes if we vary \( x \) and \( y \) independently.

• To find \( \frac{\partial v}{\partial x} \), treat \( y \) as a constant.
• To find \( \frac{\partial v}{\partial y} \), treat \( x \) as a constant.

This is essential in multivariable calculus as it provides a deeper insight into the behavior of functions with several variables. By evaluating these, we gain valuable information about the slope of the function's surface along the respective axes. This process is crucial before proceeding to further analysis like finding differentials.

Chain Rule

The chain rule in calculus is a powerful tool for taking derivatives of composed functions. When a function inside another function depends on more than one variable, the chain rule helps us differentiate it effectively.
For our function \( v = y \cos(xy) \), the chain rule applies because \( x \) and \( y \) are multiplied inside the cosine.

• First, differentiate the outer function (\( \cos \)) which turns into \( -\sin \), keeping the inner function unchanged.
• Then multiply by the derivative of the inner function (\( xy \)).
  • When \( x \) is variable, the inner derivative is \( y \).
  • When \( y \) is variable, the inner derivative is \( x \).

Utilizing the chain rule in this manner reveals the rate at which the angle \( xy \) affects the cosine output, capturing the intricate dependencies within the function.

Product Rule

The product rule is another essential concept in differential calculus, used when differentiating products of two or more functions. In our context, it's applied to the function \( v = y \cos(xy) \), which is a product of \( y \) and \( \cos(xy) \).
The product rule states: \[(fg)' = f'g + fg'\]In simple words, this means the derivative of a product is the derivative of the first function times the second, plus the first function times the derivative of the second.

• When finding \( \frac{\partial v}{\partial x} \), treat \( y \) as a constant, identify \( y \) as the first function and \( \cos(xy) \) as the second.
• For \( \frac{\partial v}{\partial y} \), treat \( x \) as constant with the same format.

Using the product rule helps capture the changes contributed by each part of the product, ensuring accuracy in calculating partial derivatives.

Differentials

Differentials highlight how small changes in variables affect changes in the function's output. In differential calculus, differentials complement the concept of derivatives by translating infinetesimal changes into a more tangible form.
For the function \( v = y \cos(xy) \), the differential, represented as \( dv \), reveals how tiny changes in \( x \) and \( y \) influence \( v \). The expression derived is:\[dv = \frac{\partial v}{\partial x}dx + \frac{\partial v}{\partial y}dy\]

• \( dx \) and \( dy \) are small changes in \( x \) and \( y \).
• \( \frac{\partial v}{\partial x} \) tells how \( v \) changes due to \( dx \).
• \( \frac{\partial v}{\partial y} \) tells how \( v \) changes due to \( dy \).

Integrating all these elements gives us a clearer representation of the function's sensitivity to changes in its variables, a practical perspective used in many fields like engineering and physics.