MCQ
જો $y = {a^x}.{b^{2x - 1}}$, તો ${{{d^2}y} \over {d{x^2}}} = . . . .$
- A${y^2}.\log a{b^2}$
- B$y.\log a{b^2}$
- C${y^2}$
- ✓$y.{(\log a{b^2})^2}$
$\frac{{dy}}{{dx}} = {a^x}{b^{2x - 1}}\log a + 2{a^x}{b^{2x - 1}}\log b$
$= {a^x}{b^{2x - 1}}(\log a + 2\log b)$
$\frac{{{d^2}y}}{{d{x^2}}} = {a^x}{b^{2x - 1}}{(\log a + 2\log b)^2}$
$ = {a^x}{b^{2x - 1}}{(\log a{b^2})^2}$$ = y{(\log a{b^2})^2}$.
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