Math Problem Statement

If \( z = \sin^{-1} \left( \frac{x}{y} \right) \), then verify that \( \frac{\partial^2 z}{\partial y \partial x} = \frac{\partial^2 z}{\partial x \partial y} \).

Solution

To verify that 2zyx=2zxy\frac{\partial^2 z}{\partial y \partial x} = \frac{\partial^2 z}{\partial x \partial y} for z=sin1(xy)z = \sin^{-1} \left( \frac{x}{y} \right), let's proceed with finding the mixed partial derivatives.

Step-by-Step Solution

  1. Given Function: z=sin1(xy)z = \sin^{-1} \left( \frac{x}{y} \right).

  2. First Partial Derivative with respect to xx: zx=11(xy)21y=1y1x2y2=yyy2x2=1y2x2\frac{\partial z}{\partial x} = \frac{1}{\sqrt{1 - \left( \frac{x}{y} \right)^2}} \cdot \frac{1}{y} = \frac{1}{y \sqrt{1 - \frac{x^2}{y^2}}} = \frac{y}{y \sqrt{y^2 - x^2}} = \frac{1}{\sqrt{y^2 - x^2}}

  3. First Partial Derivative with respect to yy: zy=11(xy)2(xy2)=xyy2x2\frac{\partial z}{\partial y} = \frac{1}{\sqrt{1 - \left( \frac{x}{y} \right)^2}} \cdot \left( -\frac{x}{y^2} \right) = -\frac{x}{y \sqrt{y^2 - x^2}}

  4. Mixed Partial Derivative 2zyx\frac{\partial^2 z}{\partial y \partial x}: y(zx)=y(1y2x2)=y(y2x2)3/2\frac{\partial}{\partial y} \left( \frac{\partial z}{\partial x} \right) = \frac{\partial}{\partial y} \left( \frac{1}{\sqrt{y^2 - x^2}} \right) = -\frac{y}{(y^2 - x^2)^{3/2}}

  5. Mixed Partial Derivative 2zxy\frac{\partial^2 z}{\partial x \partial y}: x(zy)=x(xyy2x2)=y(y2x2)3/2\frac{\partial}{\partial x} \left( \frac{\partial z}{\partial y} \right) = \frac{\partial}{\partial x} \left( -\frac{x}{y \sqrt{y^2 - x^2}} \right) = -\frac{y}{(y^2 - x^2)^{3/2}}

Since 2zyx=2zxy\frac{\partial^2 z}{\partial y \partial x} = \frac{\partial^2 z}{\partial x \partial y}, the equality is verified.

Would you like further details on each differentiation step, or do you have any questions?


Here are 5 related questions that could expand your understanding:

  1. What are mixed partial derivatives, and when are they equal?
  2. How can the chain rule be applied in partial differentiation?
  3. How does one interpret the inverse trigonometric function in the context of partial derivatives?
  4. What are some physical applications of mixed partial derivatives?
  5. Can mixed partial derivatives be unequal, and under what conditions?

Tip: For inverse trigonometric functions, always check if simplification is possible by considering the function's argument structure before applying derivative rules directly.

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Math Problem Analysis

Mathematical Concepts

Calculus
Partial Derivatives
Inverse Trigonometric Functions
Mixed Partial Derivatives

Formulas

First partial derivative of \( \sin^{-1} \left( \frac{x}{y} \right) \) with respect to \( x \): \( \frac{1}{y \sqrt{y^2 - x^2}} \)
First partial derivative of \( \sin^{-1} \left( \frac{x}{y} \right) \) with respect to \( y \): \( -\frac{x}{y \sqrt{y^2 - x^2}} \)

Theorems

Clairaut's Theorem on Equality of Mixed Partials

Suitable Grade Level

Undergraduate Calculus