MHT CET · Maths · Differentiation
If \(x \cdot \log _{\mathrm{e}}\left(\log _{\mathrm{e}} x\right)-x^2+\mathrm{y}^2=4(\mathrm{y}>0)\), then \(\frac{\mathrm{dy}}{\mathrm{d} x}\) at \(x=\mathrm{e}\) is
- A \(\frac{\mathrm{e}}{\sqrt{4+\mathrm{e}^2}}\)
- B \(\frac{2 e-1}{2 \sqrt{4+e^2}}\)
- C \(\frac{1+2 \mathrm{e}}{\sqrt{4+\mathrm{e}^2}}\)
- D \(\frac{1+2 e}{2 \sqrt{4+e^2}}\)
Answer & Solution
Correct Answer
(B) \(\frac{2 e-1}{2 \sqrt{4+e^2}}\)
Step-by-step Solution
Detailed explanation
If \(x=\mathrm{e}\): \(\mathrm{e} \cdot \log _{\mathrm{e}}\left(\log _{\mathrm{e}} \mathrm{e}\right)-\mathrm{e}^2+\mathrm{y}^2=4 \implies \mathrm{e} \cdot 0-\mathrm{e}^2+\mathrm{y}^2=4 \implies \mathrm{y}^2=4+\mathrm{e}^2 \implies \mathrm{y}=\sqrt{4+\mathrm{e}^2}\) \(\frac{\mathrm{d}}{\mathrm{d} x}\left[x \cdot \log _{\mathrm{e}}\left(\log _{\mathrm{e}} x\right)-x^2+\mathrm{y}^2\right]=\frac{\mathrm{d}}{\mathrm{d} x}[4]\)
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