MHT CET · Maths · Differentiation
If \(x^y \cdot y^x=16\), then \(\frac{d y}{d x}\) at \((2,2)\) is
- A -1
- B 0
- C 1
- D 2
Answer & Solution
Correct Answer
(A) -1
Step-by-step Solution
Detailed explanation
\(x^y \cdot y^x=16\)
Taking \(\log\) on both sides, \(\therefore \mathrm{y} \log \mathrm{x}+\mathrm{x} \log \mathrm{y}=\log 16\)
Differentiating w.r.t. \(x, \therefore \frac{y}{x}+(\log x) \frac{d y}{d x}+\left(\frac{x}{y}\right) \frac{d y}{d x}+\log y=0\)
\(\therefore\left[\log x+\frac{x}{y}\right] \frac{d y}{d x}=-\left[\frac{y}{x}+\log y\right] \)
\( \therefore \frac{d y}{d x}=\frac{-\left[\frac{y}{x}+\log y\right]}{\left[\log x+\frac{x}{y}\right]} \Rightarrow\left(\frac{d y}{d x}\right)_{(2,2)}=-\left[\frac{(1+\log 2)}{\log 2+1}\right]=-1\)
Taking \(\log\) on both sides, \(\therefore \mathrm{y} \log \mathrm{x}+\mathrm{x} \log \mathrm{y}=\log 16\)
Differentiating w.r.t. \(x, \therefore \frac{y}{x}+(\log x) \frac{d y}{d x}+\left(\frac{x}{y}\right) \frac{d y}{d x}+\log y=0\)
\(\therefore\left[\log x+\frac{x}{y}\right] \frac{d y}{d x}=-\left[\frac{y}{x}+\log y\right] \)
\( \therefore \frac{d y}{d x}=\frac{-\left[\frac{y}{x}+\log y\right]}{\left[\log x+\frac{x}{y}\right]} \Rightarrow\left(\frac{d y}{d x}\right)_{(2,2)}=-\left[\frac{(1+\log 2)}{\log 2+1}\right]=-1\)
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