AP EAMCET · Maths · Differentiation
If \(y=\frac{\log _e x}{x}\) and \(z=\log _e x\), then \(\frac{d^2 y}{d z^2}+\frac{d y}{d z}\) is equal to
- A \(e^{-z}\)
- B \(2 e^{-z}\)
- C \(z e^{-z}\)
- D \(-e^{-z}\)
Answer & Solution
Correct Answer
(D) \(-e^{-z}\)
Step-by-step Solution
Detailed explanation
Given, \(y=\frac{\log _e x}{x}\) and \(z=\log _e x\) \(\therefore \quad y=\frac{z}{e^z}\) On differentiating w.r.t. \(z\), we get \(\frac{d y}{d z}=\frac{e^z(1)-z e^z}{e^{2 z}}=\frac{1-z}{e^z}\) Again differentiating, we get…
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