If a y^4 = (x + b)^5,then frac{y cdot (frac{d | vedclass

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(C) Given the equation $a y^4 = (x + b)^5$. Taking the natural logarithm on both sides: $\ln(a) + 4 \ln(y) = 5 \ln(x + b)$. Differentiating with respe

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$\frac{1}{5}$

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(C) Given the equation $a y^4 = (x + b)^5$. Taking the natural logarithm on both sides: $\ln(a) + 4 \ln(y) = 5 \ln(x + b)$. Differentiating with respect to $x$: $\frac{4}{y} \cdot \frac{dy}{dx} = \frac{5}{x + b}$. Thus,$\frac{dy}{dx} = \frac{5y}{4(x + b)}$. Differentiating again with respect to $x$ using the quotient rule: $\frac{d^2y}{dx^2} = \frac{5}{4} \cdot \frac{(x + b) \frac{dy}{dx} - y}{(x + b)^2}$. Substitute $\frac{dy}{dx} = \frac{5y}{4(x + b)}$ into the expression: $\frac{d^2y}{dx^2} = \frac{5}{4} \cdot \frac{(x + b) \cdot \frac{5y}{4(x + b)} - y}{(x + b)^2} = \frac{5}{4} \cdot \frac{\frac{5y}{4} - y}{(x + b)^2} = \frac{5}{4} \cdot \frac{y}{4(x + b)^2} = \frac{5y}{16(x + b)^2}$. Now,calculate the ratio: $\frac{y \cdot (\frac{d^2y}{dx^2})}{(\frac{dy}{dx})^2} = \frac{y \cdot \frac{5y}{16(x + b)^2}}{(\frac{5y}{4(x + b)})^2} = \frac{\frac{5y^2}{16(x + b)^2}}{\frac{25y^2}{16(x + b)^2}} = \frac{5}{25} = \frac{1}{5}$.

If $a y^4 = (x + b)^5$,then $\frac{y \cdot (\frac{d^2 y}{d x^2})}{(\frac{d y}{d x})^2} = $

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