MHT CET · Maths · Differentiation
If \(x^y \cdot y^x=16\), then \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) at \((2,2)\) is
- A -2
- B 2
- C 1
- D -1
Answer & Solution
Correct Answer
(D) -1
Step-by-step Solution
Detailed explanation
\(x^y \cdot y^x=16\)
\(\Rightarrow y \log x+x \log y=\log 16\) [Taking log both sides]
Differentiating we get
\(\frac{\mathrm{d} y}{\mathrm{~d} x} \cdot \log x+y \cdot \frac{1}{x}+\log y+x \cdot \frac{1}{y} \cdot \frac{\mathrm{d} y}{\mathrm{~d} x}=0\)
Putting \(x=2\) and \(y=0\)
\(\begin{aligned}
& \frac{\mathrm{d} y}{\mathrm{~d} x}(1+\log 2)=-(1+\log 2) \\
& \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=-1
\end{aligned}\)
\(\Rightarrow y \log x+x \log y=\log 16\) [Taking log both sides]
Differentiating we get
\(\frac{\mathrm{d} y}{\mathrm{~d} x} \cdot \log x+y \cdot \frac{1}{x}+\log y+x \cdot \frac{1}{y} \cdot \frac{\mathrm{d} y}{\mathrm{~d} x}=0\)
Putting \(x=2\) and \(y=0\)
\(\begin{aligned}
& \frac{\mathrm{d} y}{\mathrm{~d} x}(1+\log 2)=-(1+\log 2) \\
& \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=-1
\end{aligned}\)
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