Question

If \(x^y+y^x=\mathrm{a}^{\mathrm{b}}\), then \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) at \(x=1, y=2\) is

• A. \(-2(1+\mathrm{log}2)\)
• B. \(2(1+\mathrm{log}2)\)
• C. \(2+\mathrm{log}2\)
• D. \(1+\mathrm{log}2\)

Hint:

Differentiate each term by logarithmic differentiation and substitute x = 1, y = 2.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The right side \(a^b\) is a constant, so differentiate \(x^y+y^x\) with respect to \(x\) and equate to 0.

Step 2: Differentiate the terms
\(\frac{d}{dx}x^y=x^y\left(\frac yx+\log x\cdot y'\right)\)
\(\frac{d}{dx}y^x=y^x\left(\log y+\frac{x}{y}y'\right)\).

Step 3: At (1, 2)
First term: \(1\cdot(2+0)=2\). Second term: \(2\left(\log2+\frac{y'}{2}\right)=2\log2+y'\).

Step 4: Solve
\(2+2\log2+y'=0\Rightarrow y'=-2(1+\log2)\).

Final Answer:
The derivative is \(-2(1+\log2)\). \[ \boxed{\text{(A) }-2(1+\log 2)} \]