Question

$$\text { If } z=\cos (x y), \text { show that } x \frac{\partial z}{\partial x}-y \frac{\partial z}{\partial y}=0$$.

Short answer

Expression is zero upon substitution and simplification.

Step-by-step solution

Understand the Problem

Given the function \(z = \cos(xy)\), we need to show that \(x \frac{\partial z}{\partial x} - y \frac{\partial z}{\partial y} = 0\).

Compute Partial Derivative with Respect to x

Find the partial derivative of \(z\) with respect to \(x\). Using the chain rule, $$\frac{\partial z}{\partial x} = \- \sin(xy) \cdot y $$.

Compute Partial Derivative with Respect to y

Next, find the partial derivative of \(z\) with respect to \(y\). Again, using the chain rule, $$\frac{\partial z}{\partial y} = \- \sin(xy) \cdot x $$.

Substitute and Simplify

Now substitute the partial derivatives into the original expression to check if the equation holds. $$x \frac{\partial z}{\partial x} - y \frac{\partial z}{\partial y} = x(-\sin(xy) \cdot y) - y(- \sin(xy) \cdot x)$$ Simplify this: $$= -xy \sin(xy) + yx \sin(xy)$$ $$= 0$$.

Conclusion

Since the left-hand side equals to zero, the original statement is proven: $$x\frac{\partial z}{\partial x} - y\frac{\partial z}{\partial y} = 0$$.

Key concepts

Chain Rule

The chain rule is an important technique in calculus used for finding the derivative of composite functions. In the context of partial differential equations, it helps in finding the partial derivatives when a function depends on multiple variables. For example, if we have a function \(z = \f(x, y)\), where \(z\) depends on \(x\) and \(y\), and both \(x\) and \(y\) depend on another variable \(t\), the chain rule gives us a way to calculate how \(z\) changes with respect to \(t\). This is done through the derivatives of \(z\) with respect to \(x\) and \(y\), multiplied by the derivatives of \(x\) and \(y\) with respect to \(t\). It’s useful to represent this understanding in the form:

• \frac{\text{d}z}{\text{d}t} = \frac{\frac{\text{d}z}{\text{d}x} \text{dx}}{\text{dt}} + \frac{\frac{\text{d}z}{\text{d}y} \text{dy}}{\text{dt}}

This same concept is extended to functions like \(\text{cos}(xy)\), where \(xy\) is a composite function.

Partial Derivatives

Partial derivatives measure how a multivariable function changes as one of its variables is varied while keeping the others constant. For the function \(z = \text{cos}(xy)\), we have to find the partial derivatives \frac{\frac{\text{dz}}{\text{dx}} and \frac{\frac{\text{dz}}{\text{dy}}:

• \frac{\frac{\text{dz}}{\text{dx}} = - \text{sin}(xy) \times y
• \frac{\frac{\text{dz}}{\text{dy}} = - \text{sin}(xy) \times x

In each case, we hold one of the variables constant (either \(x\) or \(y\)) and differentiate with respect to the other variable.

Trigonometric Functions

Trigonometric functions like \(\text{cos}\) and \(\text{sin}\) are fundamental in understanding oscillatory behaviors and periodic phenomena. They also have practical applications in physics and engineering. For the function \(z = \text{cos}(xy)\):

• The derivative of \(\text{cos}(u)\) with respect to \(u\) is \(-\text{sin}(u)\).
• Using this, we can find the partial derivatives needed for the exercise.

So for \(\frac{\text{d}\text{cos}(xy)}{\text{dx}}\), treat \(xy\) as a single unit and apply the chain rule: \(\frac{\text{d}\text{cos}(xy)}{\text{dx}} = -\text{sin}(xy) \times y\). Similarly, apply the chain rule for the partial derivative with respect to \(y\).

Proof Techniques

Proof techniques in mathematics give structured ways to show why certain statements are true. For the exercise, the steps to prove \(x\frac{\frac{\text{dz}}{\text{dx}} - y\frac{\frac{\text{dz}}{\text{dy}} = 0}\) involve:

• Finding the necessary partial derivatives.
• Substituting these derivatives back into the given equation.
• Simplifying the expression to show that it equals zero.

The final simplification shows that the left-hand side of the equation simplifies to zero, thus proving the original statement.