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5. continuity and differentiation — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths5. continuity and differentiation1 Mark
MCQ
If $y = {{{e^{2x}}\cos x} \over {x\sin x}},$ then ${{dy} \over {dx}} = $
  • ✓
    ${{{e^{2x}}[(2x - 1)\cot x - x\,{\rm{cose}}{{\rm{c}}^2}x]} \over {{x^2}}}$
  • B
    ${{{e^{2x}}[(2x + 1)\cot x - x\,{\rm{cose}}{{\rm{c}}^2}x]} \over {{x^2}}}$
  • C
    ${{{e^{2x}}[(2x - 1)\cot x + x\,{\rm{cose}}{{\rm{c}}^2}x]} \over {{x^2}}}$
  • D
    None of these
✓

Answer

Correct option: A.
${{{e^{2x}}[(2x - 1)\cot x - x\,{\rm{cose}}{{\rm{c}}^2}x]} \over {{x^2}}}$
a
(a) $y = \frac{{{e^{2x}}\cos x}}{{x\sin x}}$

==> $\log y = 2x + \log \cos x - \log x - \log \sin x$

$\frac{1}{y}\frac{{dy}}{{dx}} = 2 + \left( {\frac{{ - \sin x}}{{\cos x}}} \right) - \frac{1}{x} - \frac{{\cos x}}{{\sin x}}$

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

$ = \frac{{{e^{2x}}}}{{{x^2}}}[(2x - 1)\cot x - x\,{\rm{cose}}{{\rm{c}}^2}x]$.

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