If y = frac{sqrt{x}(2x + 3)^2}{sqrt{x + 1}},t | vedclass

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(A) Given $y = \frac{\sqrt{x}(2x + 3)^2}{\sqrt{x + 1}}$.Taking natural logarithm on both sides:$\ln y = \ln \left( \frac{\sqrt{x}(2x + 3)^2}{\sqrt{x +

Vedclass · Accepted Answer

$y \left[ \frac{1}{2x} + \frac{4}{2x + 3} - \frac{1}{2(x + 1)} \right]$

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(A) Given $y = \frac{\sqrt{x}(2x + 3)^2}{\sqrt{x + 1}}$.Taking natural logarithm on both sides:$\ln y = \ln \left( \frac{\sqrt{x}(2x + 3)^2}{\sqrt{x + 1}} \right)$$\ln y = \ln(\sqrt{x}) + \ln((2x + 3)^2) - \ln(\sqrt{x + 1})$$\ln y = \frac{1}{2} \ln x + 2 \ln(2x + 3) - \frac{1}{2} \ln(x + 1)$Differentiating both sides with respect to $x$:$\frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{x} + 2 \cdot \frac{1}{2x + 3} \cdot 2 - \frac{1}{2} \cdot \frac{1}{x + 1}$$\frac{1}{y} \frac{dy}{dx} = \frac{1}{2x} + \frac{4}{2x + 3} - \frac{1}{2(x + 1)}$Therefore,$\frac{dy}{dx} = y \left[ \frac{1}{2x} + \frac{4}{2x + 3} - \frac{1}{2(x + 1)} \right]$.

If $y = \frac{\sqrt{x}(2x + 3)^2}{\sqrt{x + 1}}$,then $\frac{dy}{dx} = $

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