Problem 20 - The Calculus with Analytic Geometry

Question

In Exercises 19 through 22, assume that the given equation defines \(z\) as a function of \(x\) and \(y\). Differentiate implicitly to find \(\partial z / \partial x\) and \(\partial z / \partial y\). $$ z=\left(x^{2}+y^{2}\right) \sin x z $$

Step-by-Step Solution

Answer

Implicit differentiation yields \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\).

Step 1: Differentiate with respect to x

Start by differentiating the given equation implicitly with respect to x. Apply the product rule to the term \(\left(x^{2}+y^{2}\right) \sin(xz)\). \[\frac{\partial z}{\partial x} = \frac{d}{dx}\left((x^2 + y^2) \sin(xz)\right)\]

Step 2: Apply the product rule

Applying the product rule to the right-hand side:\[\frac{\partial z}{\partial x} = \left( \frac{\partial (x^2 + y^2)}{\partial x} \right) \sin(xz) + (x^2 + y^2) \frac{\partial}{\partial x} \left( \sin(xz) \right)\]

Step 3: Differentiate each term separately

Compute each derivative separately. \[\frac{\partial (x^2 + y^2)}{\partial x} = 2x\]and \[\frac{\partial}{\partial x} \sin(xz) = \cos(xz) \left( \frac{z \partial x}{\partial x} + x \frac{\partial z}{\partial x} \right)\]

Step 4: Use the chain rule

Using the chain rule, we get: \[\frac{\partial}{\partial x} \sin(xz) = \cos(xz) \left( z + x \frac{\partial z}{\partial x} \right)\]

Step 5: Insert derivatives back into equation

Combine and insert these values back into the equation: \[\frac{\partial z}{\partial x} = 2x \sin(xz) + (x^2 + y^2) \cos(xz) \left( z + x \frac{\partial z}{\partial x} \right)\]

Step 6: Solve the equation for \( \frac{\partial z}{\partial x} \)

Rearrange the equation to isolate \( \frac{\partial z}{\partial x} \): \[\frac{\partial z}{\partial x} = 2x \sin(xz) + (x^2 + y^2) \cos(xz) (z + x \frac{\partial z}{\partial x})\]Simplify and solve for \( \frac{\partial z}{\partial x} \).

Step 7: Differentiate with respect to y

Now differentiate the original equation implicitly with respect to y: \[\frac{\partial z}{\partial y} = \frac{d}{dy} \left( (x^2 + y^2) \sin(xz) \right)\]

Step 8: Apply the product rule

Applying the product rule to the right-hand side:\[\frac{\partial z}{\partial y} = \left( \frac{\partial (x^2 + y^2)}{\partial y} \right) \sin(xz) + (x^2 + y^2) \frac{\partial}{\partial y} \left( \sin(xz) \right)\]

Step 9: Differentiate each term separately

Compute each derivative separately. \[\frac{\partial (x^2 + y^2)}{\partial y} = 2y\]and \[\frac{\partial}{\partial y} \sin(xz) = \cos(xz) \left( x \frac{\partial z}{\partial y} \right)\]

Step 10: Use the chain rule

Using the chain rule, we get: \[\frac{\partial}{\partial y} \sin(xz) = \cos(xz) \left( x \frac{\partial z}{\partial y} \right)\]

Step 11: Insert derivatives back into equation

Combine and insert these values back into the equation: \[\frac{\partial z}{\partial y} = 2y \sin(xz) + (x^2 + y^2) \cos(xz) (x \frac{\partial z}{\partial y})\]

Step 12: Solve the equation for \( \frac{\partial z}{\partial y} \)

Rearrange the equation to isolate \( \frac{\partial z}{\partial y} \): \[\frac{\partial z}{\partial y} = 2y \sin(xz) + (x^2 + y^2) \cos(xz) (x \frac{\partial z}{\partial y})\]Simplify and solve for \( \frac{\partial z}{\partial y} \).

Key Concepts

Partial Derivatives

To understand how functions change, we use derivatives. When dealing with functions of more variables, partial derivatives come into play.

Partial derivatives involve taking the derivative of a function with respect to one variable, while treating all other variables as constants. This is crucial in multivariable calculus and helps to study the behavior of complex functions.

For our problem, we have the function \(z = (x^2 + y^2) \text{sin}(xz)\). To find the partial derivatives \(\frac{\frac{dz}{dx}}\) and \(\frac{\frac{dz}{dy}}\), we differentiate the equation implicitly with respect to x and y, separately, and analyze how each variable influences the function.

Key steps include:

• Applying differentiation rules specific to the variable considered.
• Using other variables as constants during differentiation.
• Understanding that partial derivatives measure the rate of change with respect to one variable at a time, which is crucial for solving our given function.

Product Rule

When differentiating products of functions, the product rule is an essential tool. The product rule states that to differentiate a product of two functions, you multiply the first function by the derivative of the second, then add the second function multiplied by the derivative of the first.

The general formula is:

Given functions \(u\) and \(v\), the derivative of their product \(uv\) with respect to x is:

\frac{\frac{d}{dx}}(uv) = u \frac{\frac{dv}{dx}} + v \frac{\frac{du}{dx}}

For our function \(z = (x^2 + y^2) \text{sin}(xz)\), we apply the product rule when differentiating the term \((x^2 + y^2) \text{sin}(xz)\) with respect to x:

\frac{\frac{\frac{dz}{dx}}} = \frac{\frac{d}{dx}}((x^2 + y^2) \text{sin}(xz)) = ( \frac{x^2 \frac{+ y^2}{x}}) \text{sin}(xz) + ( x \frac{^2 + y^2) \frac{d}{dx}}}{( \text{sin}(xz))}

Applying each of these differentiation steps helps in breaking down the problem and obtaining the necessary derivatives.

Understanding and applying the product rule is fundamental in implicit differentiation, as it handles expressions where variables are multiplied together.

Chain Rule

The chain rule is another critical technique in calculus, particularly when dealing with composite functions. This rule allows us to differentiate a function that contains another function inside it.

The chain rule formula looks like this:

• If \(y = g(f(x))\), then \(\frac{\frac{dy}{dx}} = \frac{\frac{dg}}{du} \frac{du}{dx}\) where \(u = f(x)\).

For our function, we applied the chain rule to differentiate terms like \(\text{sin}(xz)\). Since \(xz\) is a function of both x and z, we need to consider both parts:

\frac{\frac{d}}{dx} (\text{sin}(xz)) = \text{cos}(xz) \frac{\frac{d}}{dx} (xz) = \text{cos}(xz) (z + x \frac{\frac{dz}{dx}})

The same principle applies when differentiating \(\text{sin}(xz)\) with respect to y:

\frac{\frac{d}}{dy} (\text{sin}(xz)) = \text{cos}(xz) \frac{\frac{d}}{dy} (xz) = \text{cos}(xz) (x \frac{\frac{dz}{dy}})

This allows us to handle the complexity of nested functions and understand how the inner function's change affects the outer function.

Mastering the chain rule is essential for implicit differentiation, enabling us to break down complex relationships within the function.