If y = a^x cdot b^{2x - 1},then frac{d^2y}{dx | vedclass

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(D) Given $y = a^x \cdot b^{2x - 1}$.Taking the natural logarithm on both sides: $\ln y = x \ln a + (2x - 1) \ln b$.Differentiating with respect to $x

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$y \cdot (\log ab^2)^2$

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(D) Given $y = a^x \cdot b^{2x - 1}$.Taking the natural logarithm on both sides: $\ln y = x \ln a + (2x - 1) \ln b$.Differentiating with respect to $x$: $\frac{1}{y} \frac{dy}{dx} = \ln a + 2 \ln b$.Thus,$\frac{dy}{dx} = y(\ln a + \ln b^2) = y \ln(ab^2)$.Differentiating again with respect to $x$: $\frac{d^2y}{dx^2} = \frac{dy}{dx} \ln(ab^2)$.Substituting the value of $\frac{dy}{dx}$: $\frac{d^2y}{dx^2} = [y \ln(ab^2)] \cdot \ln(ab^2) = y(\ln(ab^2))^2$.

If $y = a^x \cdot b^{2x - 1}$,then $\frac{d^2y}{dx^2}$ is

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