Math Problem Statement

Find the first-order partial derivatives and other relations involving multivariable functions such as u = tan^(-1)((x^2 + y^2)/(x + y)) and u = cos^(-1)(x/y). Prove identities involving second and third partial derivatives.

Solution

The image contains several partial derivative problems. Let's break them down and work through the first problem in detail:

Problem 1:

Find the first-order partial derivatives of the following:

(i) u=tan⁡−1(x2+y2x+y)u = \tan^{-1} \left( \frac{x^2 + y^2}{x + y} \right)

(ii) u=cos⁡−1(xy)u = \cos^{-1} \left( \frac{x}{y} \right)

Solution for (i):

Given: u=tan⁡−1(x2+y2x+y)u = \tan^{-1} \left( \frac{x^2 + y^2}{x + y} \right)

To find the partial derivatives ∂u∂x\frac{\partial u}{\partial x} and ∂u∂y\frac{\partial u}{\partial y}, we use the chain rule. For tan⁡−1(v)\tan^{-1}(v), we know: ddvtan⁡−1(v)=11+v2\frac{d}{dv} \tan^{-1}(v) = \frac{1}{1 + v^2}

Let v=x2+y2x+yv = \frac{x^2 + y^2}{x + y}. Then, we need to compute the partial derivatives of vv with respect to xx and yy.

Partial derivative with respect to xx:

For v=x2+y2x+yv = \frac{x^2 + y^2}{x + y}, apply the quotient rule: ∂v∂x=(2x)(x+y)−(x2+y2)(1)(x+y)2\frac{\partial v}{\partial x} = \frac{(2x)(x + y) - (x^2 + y^2)(1)}{(x + y)^2} Simplify:

= \frac{2x^2 + 2xy - x^2 - y^2}{(x + y)^2} = \frac{x^2 + 2xy - y^2}{(x + y)^2}$$ Now apply the chain rule: $$\frac{\partial u}{\partial x} = \frac{1}{1 + v^2} \cdot \frac{\partial v}{\partial x}$$ ##### Partial derivative with respect to $$y$$: Similarly, for $$v = \frac{x^2 + y^2}{x + y}$$, apply the quotient rule with respect to $$y$$: $$\frac{\partial v}{\partial y} = \frac{(2y)(x + y) - (x^2 + y^2)(1)}{(x + y)^2}$$ Simplify: $$\frac{\partial v}{\partial y} = \frac{2y(x + y) - (x^2 + y^2)}{(x + y)^2} = \frac{2xy + 2y^2 - x^2 - y^2}{(x + y)^2} = \frac{-x^2 + 2xy + y^2}{(x + y)^2}$$ Thus, the chain rule gives: $$\frac{\partial u}{\partial y} = \frac{1}{1 + v^2} \cdot \frac{\partial v}{\partial y}$$ --- #### Solution for (ii): Given: $$u = \cos^{-1} \left( \frac{x}{y} \right)$$ We use the chain rule here as well. For $$\cos^{-1}(v)$$, we know: $$\frac{d}{dv} \cos^{-1}(v) = -\frac{1}{\sqrt{1 - v^2}}$$ Let $$v = \frac{x}{y}$$. Then we need to compute the partial derivatives of $$v$$ with respect to $$x$$ and $$y$$. ##### Partial derivative with respect to $$x$$: For $$v = \frac{x}{y}$$, the derivative with respect to $$x$$ is: $$\frac{\partial v}{\partial x} = \frac{1}{y}$$ Thus, using the chain rule: $$\frac{\partial u}{\partial x} = -\frac{1}{\sqrt{1 - \left( \frac{x}{y} \right)^2}} \cdot \frac{1}{y}$$ ##### Partial derivative with respect to $$y$$: For $$v = \frac{x}{y}$$, the derivative with respect to $$y$$ is: $$\frac{\partial v}{\partial y} = -\frac{x}{y^2}$$ Thus, using the chain rule: $$\frac{\partial u}{\partial y} = -\frac{1}{\sqrt{1 - \left( \frac{x}{y} \right)^2}} \cdot \left( -\frac{x}{y^2} \right)$$ --- Do you need further steps or clarification on these partial derivative problems? Let me know if you'd like me to solve additional problems from this list or go into more detail on any steps. Here are 5 related questions: 1. How does the chain rule apply to multivariable functions in general? 2. How would you simplify the second derivative for $$u = \tan^{-1} \left( \frac{x^2 + y^2}{x + y} \right)$$? 3. How do you interpret the geometrical meaning of partial derivatives in two variables? 4. What happens when you mix derivatives like $$\frac{\partial^2 u}{\partial x \partial y}$$? 5. How does implicit differentiation work in cases involving inverse trigonometric functions? **Tip**: Always double-check that you're applying the chain rule correctly when differentiating composite functions like trigonometric and inverse trigonometric functions.

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Math Problem Analysis

Mathematical Concepts

Partial Derivatives
Chain Rule
Quotient Rule
Inverse Trigonometric Functions
Multivariable Calculus

Formulas

Partial derivative of tan^(-1)(v): d/dv [tan^(-1)(v)] = 1/(1 + v^2)
Partial derivative of cos^(-1)(v): d/dv [cos^(-1)(v)] = -1/sqrt(1 - v^2)
Quotient rule for derivatives: d/dx [f(x)/g(x)] = (f'(x)g(x) - f(x)g'(x))/(g(x)^2)

Theorems

Multivariable Chain Rule
Inverse Trigonometric Differentiation

Suitable Grade Level

Grades 11-12, College-level Calculus

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