If y = e^x x x^2 , then dy dx = - Vidyadip

MCQ

If $y = {{{e^x}\log x} \over {{x^2}}}$, then ${{dy} \over {dx}} = $

A
${{{e^x}[1 + (x + 2)\log x]} \over {{x^3}}}$
B
${{{e^x}[1 - (x - 2)\log x]} \over {{x^4}}}$
C
${{{e^x}[1 - (x - 2)\log x]} \over {{x^3}}}$
✓
${{{e^x}[1 + (x - 2)\log x]} \over {{x^3}}}$

✓

Answer

Correct option: D.

${{{e^x}[1 + (x - 2)\log x]} \over {{x^3}}}$

d
(d) $\frac{{dy}}{{dx}} = - 2{x^{ - 3}}{e^x}\log x + {x^{ - 2}}\left( {{e^x}\log x + \frac{{{e^x}}}{x}} \right)$

$ = {e^x}\left[ {\frac{{1 + (x - 2)\log x}}{{{x^3}}}} \right]$

Aliter: Taking $\log $, $\log y = x + \log \log x - 2\log x$

==> $\frac{1}{y}\frac{{dy}}{{dx}} = 1 + \frac{1}{{x\log x}} - \frac{2}{x}$

==> $\frac{{dy}}{{dx}} = \frac{{{e^x}\log x}}{{{x^2}}}$ $\left[ {\frac{{x\log x + 1 - 2\log x}}{{x\log x}}} \right]$

$ = \frac{{{e^x}[(x - 2)\log x + 1]}}{{{x^3}}}$.

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