TS EAMCET · Maths · Differentiation
If \(x>0, x^y=e^{x-y}\), then \(\frac{d y}{d x}\) is equal to
- A \(\frac{1}{(1+\log x)^2}\)
- B \(\frac{\log x}{(1+\log x)^2}\)
- C \(\left(\frac{\log x}{1+\log x}\right)^2\)
- D \(\frac{(\log x)^2}{1+\log x}\)
Answer & Solution
Correct Answer
(B) \(\frac{\log x}{(1+\log x)^2}\)
Step-by-step Solution
Detailed explanation
\(y \ln x = x - y\) \(y(1+\ln x) = x\) \(y = \frac{x}{1+\ln x}\) \(\frac{d y}{d x} = \frac{(1+\ln x) \cdot 1 - x \cdot \frac{1}{x}}{(1+\ln x)^2}\) \(\frac{d y}{d x} = \frac{1+\ln x - 1}{(1+\ln x)^2}\) \(\frac{d y}{d x} = \frac{\ln x}{(1+\ln x)^2}\)
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