If y=(x-1)^{2}(x-2)^{3}(x-3)^{5},then frac{dy | vedclass

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(D) Given $y=(x-1)^{2}(x-2)^{3}(x-3)^{5}$.Taking the natural logarithm on both sides:$\log y = \log [(x-1)^{2}(x-2)^{3}(x-3)^{5}]$Using logarithmic pr

Vedclass · Accepted Answer

$516$

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(D) Given $y=(x-1)^{2}(x-2)^{3}(x-3)^{5}$.Taking the natural logarithm on both sides:$\log y = \log [(x-1)^{2}(x-2)^{3}(x-3)^{5}]$Using logarithmic properties,$\log y = 2 \log (x-1) + 3 \log (x-2) + 5 \log (x-3)$.Differentiating both sides with respect to $x$:$\frac{1}{y} \frac{dy}{dx} = \frac{2}{x-1} + \frac{3}{x-2} + \frac{5}{x-3}$.Therefore,$\frac{dy}{dx} = y \left[ \frac{2}{x-1} + \frac{3}{x-2} + \frac{5}{x-3} ight]$.Substituting $x=4$:$y(4) = (4-1)^{2}(4-2)^{3}(4-3)^{5} = 3^{2} 	imes 2^{3} 	imes 1^{5} = 9 	imes 8 	imes 1 = 72$.$\left( \frac{dy}{dx} ight)_{x=4} = 72 \left[ \frac{2}{4-1} + \frac{3}{4-2} + \frac{5}{4-3} ight] = 72 \left[ \frac{2}{3} + \frac{3}{2} + 5 ight]$.$= 72 \left[ \frac{4 + 9 + 30}{6} ight] = 72 	imes \frac{43}{6} = 12 	imes 43 = 516$.

If $y=(x-1)^{2}(x-2)^{3}(x-3)^{5}$,then $\frac{dy}{dx}$ at $x=4$ is equal to

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