AP 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
\(x^y = e^{x-y}\) \(\log(x^y) = \log(e^{x-y})\) \(y \log x = x - y\) \(y(\log x + 1) = x\) \(y = \frac{x}{1 + \log x}\) \(\frac{dy}{dx} = \frac{(1)(1 + \log x) - x(\frac{1}{x})}{(1 + \log x)^2}\) \(\frac{dy}{dx} = \frac{1 + \log x - 1}{(1 + \log x)^2}\)…
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