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:
Find dy/dx at (2,2) for the implicit equation x^y · y^x = 16 using logarithmic differentiation.

Step 2: Key Formula or Approach:
Take ln of both sides, use log properties ln(A·B)=lnA+lnB and ln(A^B)=BlnA, then differentiate implicitly with the product rule.

Step 3: Detailed Explanation:
ln(x^y·y^x) = y ln x + x ln y = ln 16. Differentiating: (y/x + ln x·dy/dx) + (x/y·dy/dx + ln y) = 0. At (2,2): (1+ln2)dy/dx = –(1+ln2) → dy/dx = –1.

Step 4: Final Answer:
The derivative at (2,2) is –1, matching option (A).