Question

If \(x^{y}=y^{x}\), find \(d y / d x\) at \((2,4)\).

Short answer

1

Step-by-step solution

- Start with the given equation

The given equation is: \[ x^y = y^x \]

- Apply the natural logarithm

Take the natural logarithm on both sides of the equation: \[ \text{ln}(x^y) = \text{ln}(y^x) \]

- Use logarithm properties

Apply the property of logarithms, \( \text{ln}(a^b) = b \text{ln}(a) \): \[ y \text{ln}(x) = x \text{ln}(y) \]

- Differentiate implicitly with respect to x

Differentiate both sides of the equation implicitly with respect to \( x \): \[ \frac{d}{dx} [ y \text{ln}(x) ] = \frac{d}{dx} [ x \text{ln}(y) ] \]Using the product rule \( (uv)' = u'v + uv' \), this becomes: \[ y \frac{1}{x} + \text{ln}(x) \frac{dy}{dx} = \text{ln}(y) + x \frac{1}{y} \frac{dy}{dx} \]

- Solve for \( \frac{dy}{dx} \)

Rearrange the equation to isolate \( \frac{dy}{dx} \): \[ \text{ln}(x) \frac{dy}{dx} - x \frac{1}{y} \frac{dy}{dx} = \text{ln}(y) - \frac{y}{x} \]Factor out \( \frac{dy}{dx} \): \[ \frac{dy}{dx} \bigg ( \text{ln}(x) - \frac{x}{y} \bigg ) = \text{ln}(y) - \frac{y}{x} \]So: \[ \frac{dy}{dx} = \frac{\text{ln}(y) - \frac{y}{x}}{\text{ln}(x) - \frac{x}{y}} \]

- Evaluate at (2,4)

Substitute \( x = 2 \) and \( y = 4 \) into the equation: \[ \frac{dy}{dx} = \frac{\text{ln}(4) - \frac{4}{2}}{\text{ln}(2) - \frac{2}{4}} = \frac{\text{ln}(4) - 2}{\text{ln}(2) - \frac{1}{2}} \]Using \( \text{ln}(4) = 2 \text{ln}(2) \), this becomes: \[ \frac{dy}{dx} = \frac{2 \text{ln}(2) - 2}{\text{ln}(2) - \frac{1}{2}} \]

- Simplify

Simplify the expression: \[ \frac{dy}{dx} = \frac{2 (\text{ln}(2) - 1)}{\text{ln}(2) - \frac{1}{2}} \]Divide the numerator and denominator by \( \text{ln}(2) \): \[ \frac{dy}{dx} = \frac{2 (1 - \frac{1}{\text{ln}(2)})}{1 - \frac{1}{2 \text{ln}(2)}} \]

Key concepts

Natural Logarithms

Natural logarithms are logarithms with base e (approximately 2.71828). They are commonly written as \text{ln}(x). Natural logarithms have unique properties that simplify complex equations, especially in calculus.
In our exercise, taking the natural logarithm of both sides of the equation converts the exponential form into a product, making it easier to differentiate.
For example, applying \text{ln} to both sides of our equation gives us:
\[ \text{ln}(x^y) = \text{ln}(y^x) \ y \text{ln}(x) = x \text{ln}(y) \]
This transformation is crucial because it sets up the equation for differentiation using known rules.

Product Rule

The product rule is a differentiation rule used when differentiating the product of two functions. It states:
\[ (uv)' = u'v + uv' \]
This rule is useful when we have a product of two different functions dependent on the same variable.
In our exercise, we differentiate \text{y ln(x)} and \text{x ln(y)} with respect to x using the product rule:
\[ \frac{d}{dx} [ y \text{ln}(x) ] = y \frac{d}{dx} [ \text{ln}(x) ] + \text{ln}(x) \frac{d}{dx} [ y ] \]
\[ \frac{d}{dx} [ y \text{ln}(x) ] = y \frac{1}{x} + \text{ln}(x) \frac{dy}{dx} \]
Similarly, for the right-hand side:
\[ \frac{d}{dx} [ x \text{ln}(y) ] = \text{ln}(y) \frac{d}{dx} [ x ] + x \frac{d}{dx} [ \text{ln}(y) ] \]
\[ = \text{ln}(y) \times 1 + x \frac{1}{y} \frac{dy}{dx} \]
This product rule application simplifies our equation, helping us find the derivative.

Derivative Evaluation

Evaluating a derivative involves solving for \frac{dy}{dx} and most often requires isolation of \frac{dy}{dx} on one side of the equation. We use algebraic manipulation to isolate \frac{dy}{dx}.
In this exercise, our rearranged equation is:
\[ \text{ln}(x) \frac{dy}{dx} - x \frac{1}{y} \frac{dy}{dx} = \text{ln}(y) - \frac{y}{x} \]
Factoring out \frac{dy}{dx} on the left-hand side, we obtain:
\[ \frac{dy}{dx} \bigg ( \text{ln}(x) - \frac{x}{y} \bigg ) = \text{ln}(y) - \frac{y}{x} \]
Finally, we solve for \frac{dy}{dx}:
\[ \frac{dy}{dx} = \frac{\text{ln}(y) - \frac{y}{x}}{\text{ln}(x) - \frac{x}{y}} \]
Plugging in the given point (2, 4), we continue by substituting \text{x=2} and \text{y=4}:
\[ \frac{dy}{dx} = \frac{2 \text{ln}(2) - 2}{\text{ln}(2) - \frac{1}{2}} \],
Finally simplifying the fractional expression, we get the evaluated derivative at the point (2, 4).