If y=(x+3)^2 cdot(x+4)^3 cdot(x+5)^4,then,the | vedclass

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(D) Given $y = (x+3)^2 \cdot (x+4)^3 \cdot (x+5)^4$. Taking the natural logarithm on both sides: $\ln(y) = 2 \ln(x+3) + 3 \ln(x+4) + 4 \ln(x+5)$. Diff

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$y \sum_{i=2}^4 \left( \frac{i}{x+i+1} \right)$

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(D) Given $y = (x+3)^2 \cdot (x+4)^3 \cdot (x+5)^4$. Taking the natural logarithm on both sides: $\ln(y) = 2 \ln(x+3) + 3 \ln(x+4) + 4 \ln(x+5)$. Differentiating both sides with respect to $x$: $\frac{1}{y} \frac{dy}{dx} = \frac{2}{x+3} + \frac{3}{x+4} + \frac{4}{x+5}$. Thus,$\frac{dy}{dx} = y \left( \frac{2}{x+3} + \frac{3}{x+4} + \frac{4}{x+5} \right)$. We can express the sum as $\sum_{i=2}^4 \frac{i}{x+i+1}$. For $i=2$: $\frac{2}{x+2+1} = \frac{2}{x+3}$. For $i=3$: $\frac{3}{x+3+1} = \frac{3}{x+4}$. For $i=4$: $\frac{4}{x+4+1} = \frac{4}{x+5}$. Therefore,the derivative is $y \sum_{i=2}^4 \left( \frac{i}{x+i+1} \right)$,which matches option $D$.

If $y=(x+3)^2 \cdot(x+4)^3 \cdot(x+5)^4$,then,the first order derivative of $y$ with respect to $x$ is . . . . . . .

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