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STD 12 - 5. continuity and differentiation — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 5. continuity and differentiation4 Marks
MCQ
If $y = {a^x}.{b^{2x - 1}}$, then ${{{d^2}y} \over {d{x^2}}}$ is
  • A
    ${y^2}.\log a{b^2}$
  • B
    $y.\log a{b^2}$
  • C
    ${y^2}$
  • ✓
    $y.{(\log a{b^2})^2}$
✓

Answer

Correct option: D.
$y.{(\log a{b^2})^2}$
d
(d) $y = {a^x}{b^{2x - 1}}$

$\frac{{dy}}{{dx}} = {a^x}{b^{2x - 1}}\log a + 2{a^x}{b^{2x - 1}}\log b$

$= {a^x}{b^{2x - 1}}(\log a + 2\log b)$

$\frac{{{d^2}y}}{{d{x^2}}} = {a^x}{b^{2x - 1}}{(\log a + 2\log b)^2}$

$ = {a^x}{b^{2x - 1}}{(\log a{b^2})^2}$$ = y{(\log a{b^2})^2}$.

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