If y = sqrt{frac{x^4 sqrt{3x-5}}{(x^2-3)(2x-3 | vedclass

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(D) Given $y = \sqrt{\frac{x^4 \sqrt{3x-5}}{(x^2-3)(2x-3)}}$.At $x = 2$,we calculate the value of $y$:$y = \sqrt{\frac{2^4 \sqrt{3(2)-5}}{(2^2-3)(2(2)

Vedclass · Accepted Answer

$-5$

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(D) Given $y = \sqrt{\frac{x^4 \sqrt{3x-5}}{(x^2-3)(2x-3)}}$.At $x = 2$,we calculate the value of $y$:$y = \sqrt{\frac{2^4 \sqrt{3(2)-5}}{(2^2-3)(2(2)-3)}} = \sqrt{\frac{16 \sqrt{1}}{(4-3)(4-3)}} = \sqrt{\frac{16}{1 	imes 1}} = \sqrt{16} = 4$.Taking the natural logarithm on both sides:$\ln(y) = \frac{1}{2} [4 \ln(x) + \frac{1}{2} \ln(3x-5) - \ln(x^2-3) - \ln(2x-3)]$.Differentiating with respect to $x$:$\frac{1}{y} \frac{dy}{dx} = \frac{1}{2} [\frac{4}{x} + \frac{1}{2} \cdot \frac{3}{3x-5} - \frac{2x}{x^2-3} - \frac{2}{2x-3}]$.Substitute $x = 2$ and $y = 4$:$\frac{1}{4} \left(\frac{dy}{dx}ight)_{x=2} = \frac{1}{2} [\frac{4}{2} + \frac{3}{2(1)} - \frac{4}{4-3} - \frac{2}{4-3}] = \frac{1}{2} [2 + 1.5 - 4 - 2] = \frac{1}{2} [-2.5] = -1.25$.Therefore,$\left(\frac{dy}{dx}ight)_{x=2} = 4 	imes (-1.25) = -5$.

If $y = \sqrt{\frac{x^4 \sqrt{3x-5}}{(x^2-3)(2x-3)}}$,then $\left(\frac{dy}{dx}\right)_{x=2} = $

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