Question

If $x^y \cdot y^x = 16$, then $\frac{dy}{dx}$ at $(2,2)$ is

• A. $-1$
• B. 0
• C. 1
• D. 2

Hint:

Notice the perfect symmetry of the expressions in the equation: $x^y \cdot y^x = 16$. Whenever an implicit relation is completely symmetric with respect to $x$ and $y$, the value of the derivative at any point lying on the line of symmetry $y = x$ (such as $(2,2)$) will always be exactly equal to $-1$!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given an implicit equation $x^y \cdot y^x = 16$ involving variable exponents. We need to compute the numerical value of the first derivative $\frac{dy}{dx}$ specifically evaluated at the coordinate point $(2,2)$.

Step 2: Key Formula or Approach:
For equations with variable bases and variable exponents, we use the method of logarithmic differentiation. Taking the natural logarithm on both sides allows us to apply standard log expansion properties: $$\ln(A \cdot B) = \ln A + \ln B \quad \text{and} \quad \ln(A^B) = B\ln A$$ We then differentiate implicitly with respect to $x$ using the product rule.

Step 3: Detailed Explanation:
Take the natural logarithm on both sides of the given equation: $$\ln\left(x^y \cdot y^x\right) = \ln(16)$$ Using log properties, rewrite this as: $$y\ln x + x\ln y = \ln(16)$$ Now, differentiate both sides with respect to $x$. Apply the product rule $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$ to both terms on the left-hand side, keeping in mind that $\frac{d}{dx}(\ln y) = \frac{1}{y}\frac{dy}{dx}$: $$\left[ y \cdot \frac{1}{x} + \ln x \cdot \frac{dy}{dx} \right] + \left[ x \cdot \frac{1}{y}\frac{dy}{dx} + \ln y \cdot 1 \right] = 0$$ $$\frac{y}{x} + \ln x \frac{dy}{dx} + \frac{x}{y}\frac{dy}{dx} + \ln y = 0$$ Group the terms containing $\frac{dy}{dx}$ together: $$\left(\ln x + \frac{x}{y}\right)\frac{dy}{dx} = -\left(\frac{y}{x} + \ln y\right)$$ To find the value of $\frac{dy}{dx}$ at the point $(2,2)$, substitute $x = 2$ and $y = 2$ directly into this differentiated expression: $$\left(\ln 2 + \frac{2}{2}\right)\frac{dy}{dx} = -\left(\frac{2}{2} + \ln 2\right)$$ $$\left(\ln 2 + 1\right)\frac{dy}{dx} = - \left(1 + \ln 2\right)$$ Since $(\ln 2 + 1)$ is a non-zero common factor on both sides, we can cancel it out completely: $$\frac{dy}{dx} = -1$$

Step 4: Final Answer:
The value of the derivative at $(2,2)$ is $-1$, which corresponds to option (A).