If y = a^x . b^ 2x - 1 , then d^2 y d x^2 is - Vidyadip

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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