# Partial Derivatives with Inverse Trig Functions

• issisoccer10
In summary, the Cauchy-Riemann equations can be satisfied by u(x,y) = ln(x^{2} + y^{2}) and v(x,y) = 2tan^{-1} (y/x) by taking the partial derivatives of the single variable differentials and equating them to each other. After simplifying, the left and right sides are equal and the Cauchy-Riemann equations are satisfied.
issisoccer10
[SOLVED] Partial Derivatives with Inverse Trig Functions

## Homework Statement

Show that u(x,y) and v(x,y) satisfy the Cauchy-Riemann equations...

$$\frac{\partial u}{\partial x}$$ = $$\frac{\partial v}{\partial y}$$

given that u = ln(x$$^{2}$$ + y$$^{2}$$)
and that v = 2tan$$^{-1}$$ (y/x)

## Homework Equations

single variable differentials of equations should be in the form of..

$$\frac{d}{dx}$$ ln (u) = $$\frac{1}{u}$$ $$\frac{du}{dx}$$

$$\frac{d}{dx}$$ tan$$^{-1}$$ u = $$\frac{1}{1+u^{2}}$$ $$\frac{du}{dx}$$

## The Attempt at a Solution

I would think that the partial derivatives of the above equations would lead from the single differentials, but instead of $$\frac{du}{dx}$$ there would be $$\frac{\partial u}{\partial x}$$ or $$\frac{\partial u}{\partial y}$$ depending on which the question called for. With this reasoning, I thought the left side of the Cauchy-Riemann equation should be..

$$\frac{2x}{x^{2} + y ^{2}}$$

and the right side would be...

$$\frac{2}{1 + (y/x)^{2}}$$ $$\frac{1}{x}$$

However, these don't appear to be equal... unless I can't see the simple algebra. Any help would be greatly appreciated..

issisoccer10 said:
$$\frac{2x}{x^{2} + y ^{2}}$$

and the right side would be...

$$\frac{2}{1 + (y/x)^{2}}$$ $$\frac{1}{x}$$
$$\frac{2}{1+\left(\frac{y}{x}\right)^2} \times \frac{1}{x}$$
Distribute through and multply by 1 ($\frac{x}{x]$)...
$$\frac{2}{x+\frac{y^2}{x}} \times \frac{x}{x}$$
$$\frac{2x}{x^2+y^2}$$

in the denominator of your responce you have the term y^{2}/x. Shouldn't the term be y^{2}/x^{2}?

$$\frac{1}{1+\left(\frac{y}{x}\right)^2} \times \frac{1}{x}$$
$$\frac{1}{\left(1+\left(\frac{y}{x}\right)^2\right) x}$$
$$\frac{1}{x + x \times \frac{y^2}{x^2}}$$
$$\frac{1}{x + \frac{y^2}{x}}$$

wow... i feel pathetic... thanks again.

## What are partial derivatives?

Partial derivatives are a type of derivative used in multivariable calculus to calculate the rate of change of a function with respect to one of its independent variables, while holding all other variables constant.

## How are inverse trig functions used in partial derivatives?

Inverse trig functions, such as arcsine, arccosine, and arctangent, are used in partial derivatives to find the rate of change of a function that involves trigonometric functions. They are used to "undo" the trigonometric function and find the original angle that produced a specific output.

## What is the chain rule for partial derivatives?

The chain rule for partial derivatives is used when taking the derivative of a function that is composed of two or more functions. It states that the partial derivative of a composite function is equal to the product of the partial derivative of the outer function and the derivative of the inner function.

## Can inverse trig functions be differentiated using the power rule?

No, inverse trig functions cannot be differentiated using the power rule. They require the use of the chain rule or other derivative rules specific to trigonometric functions.

## How can partial derivatives with inverse trig functions be applied in real-world situations?

Partial derivatives with inverse trig functions are commonly used in physics and engineering to calculate rates of change in circular motion, such as the acceleration of an object moving in a circular path. They are also used in economics to determine the marginal utility or rate of change in utility with respect to a change in a variable.

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