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

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(C) Given $y = a^x \cdot b^{2x-1}$.We can rewrite this as $y = a^x \cdot (b^2)^x \cdot b^{-1} = \frac{1}{b} (a b^2)^x$.Taking the natural logarithm on

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$y(\log(a b^2))^2$

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(C) Given $y = a^x \cdot b^{2x-1}$.We can rewrite this as $y = a^x \cdot (b^2)^x \cdot b^{-1} = \frac{1}{b} (a b^2)^x$.Taking the natural logarithm on both sides: $\log y = \log \left( \frac{1}{b} (a b^2)^x \right) = \log(1) - \log(b) + x \log(a b^2) = -\log b + x \log(a b^2)$.Differentiating with respect to $x$: $\frac{1}{y} \frac{dy}{dx} = \log(a b^2)$.So,$\frac{dy}{dx} = y \log(a b^2)$.Differentiating again with respect to $x$: $\frac{d^2 y}{d x^2} = \frac{dy}{dx} \log(a b^2)$.Substituting $\frac{dy}{dx} = y \log(a b^2)$,we get $\frac{d^2 y}{d x^2} = (y \log(a b^2)) \cdot \log(a b^2) = y(\log(a b^2))^2$.

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

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