Math Problem Statement

derivative of x^y=y^x

Solution

To find the derivative of the equation xy=yxx^y = y^x, we can proceed with implicit differentiation. Here's the step-by-step approach:

Step 1: Take the natural logarithm of both sides

First, take the natural logarithm of both sides to simplify the equation:

ln(xy)=ln(yx)\ln(x^y) = \ln(y^x)

Using the logarithmic identity ln(ab)=bln(a)\ln(a^b) = b \ln(a), this becomes:

yln(x)=xln(y)y \ln(x) = x \ln(y)

Step 2: Differentiate implicitly

Now, differentiate both sides with respect to xx, keeping in mind that yy is a function of xx (i.e., y=y(x)y = y(x)).

  • Left side: Use the product rule to differentiate yln(x)y \ln(x).

ddx(yln(x))=dydxln(x)+y1x\frac{d}{dx} \left( y \ln(x) \right) = \frac{dy}{dx} \ln(x) + y \frac{1}{x}

  • Right side: Use the product rule to differentiate xln(y)x \ln(y).

ddx(xln(y))=ln(y)+x1ydydx\frac{d}{dx} \left( x \ln(y) \right) = \ln(y) + x \frac{1}{y} \frac{dy}{dx}

Step 3: Set up the equation

Now, equate the derivatives from both sides:

dydxln(x)+yx=ln(y)+xydydx\frac{dy}{dx} \ln(x) + \frac{y}{x} = \ln(y) + \frac{x}{y} \frac{dy}{dx}

Step 4: Solve for dydx\frac{dy}{dx}

Move the terms involving dydx\frac{dy}{dx} to one side:

dydxln(x)xydydx=ln(y)yx\frac{dy}{dx} \ln(x) - \frac{x}{y} \frac{dy}{dx} = \ln(y) - \frac{y}{x}

Factor out dydx\frac{dy}{dx} on the left-hand side:

dydx(ln(x)xy)=ln(y)yx\frac{dy}{dx} \left( \ln(x) - \frac{x}{y} \right) = \ln(y) - \frac{y}{x}

Now solve for dydx\frac{dy}{dx}:

dydx=ln(y)yxln(x)xy\frac{dy}{dx} = \frac{\ln(y) - \frac{y}{x}}{\ln(x) - \frac{x}{y}}

Final Answer:

The derivative of the equation xy=yxx^y = y^x with respect to xx is:

dydx=ln(y)yxln(x)xy\frac{dy}{dx} = \frac{\ln(y) - \frac{y}{x}}{\ln(x) - \frac{x}{y}}

Let me know if you want further clarification or details on any part!


Related Questions:

  1. How can we apply implicit differentiation to equations involving both xx and yy?
  2. What are some special cases for the equation xy=yxx^y = y^x when x=yx = y?
  3. How would you solve for yy explicitly if possible?
  4. What is the geometric interpretation of this equation and its derivative?
  5. How does the derivative behave when xx and yy approach certain values?

Tip: When differentiating equations like xy=yxx^y = y^x, taking the logarithm of both sides often simplifies the process significantly!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Implicit Differentiation
Exponential Functions
Logarithms

Formulas

ln(a^b) = b * ln(a)
Product Rule: d/dx (u*v) = u' * v + u * v'
dy/dx = (ln(y) - (y/x)) / (ln(x) - (x/y))

Theorems

Implicit Differentiation Theorem
Logarithmic Properties

Suitable Grade Level

Grades 11-12