\begin{tabular}{l|l} Find \( \frac{d^{2} y}{d x^{2}} \) in terms of \( x \) and \( y \) \\ \( y^{4}=x^{5} \) \\ \( \frac{d^{2} y}{d x^{2}}=\square \)\end{tabular}

To find \( \frac{d^{2} y}{dx^{2}} \) in terms of \( x \) and \( y \), we first need to differentiate \( y^4 = x^5 \) with respect to \( x \). 1. Differentiate both sides with respect to \( x \): \[ \frac{d}{dx}(y^4) = \frac{d}{dx}(x^5) \] Using the chain rule, we get: \[ 4y^3 \frac{dy}{dx} = 5x^4 \] Then, solving for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{5x^4}{4y^3} \] 2. Now, differentiate \( \frac{dy}{dx} \) to find \( \frac{d^2y}{dx^2} \): We apply the quotient rule: \[ \frac{d^2y}{dx^2} = \frac{(4y^3)(20x^3) - (5x^4)(12y^2 \frac{dy}{dx})}{(4y^3)^2} \] We have already found \( \frac{dy}{dx} \) earlier, so now substituting \( \frac{dy}{dx} \): \[ = \frac{80xy^3 - 60x^4 \frac{5x^4}{4y^3}}{16y^6} \] Finally, simplifying further results in: \[ \frac{d^2y}{dx^2} = \frac{80xy^3 - \frac{75x^8}{4y^3}}{16y^6} \] So the complete expression for \( \frac{d^2y}{dx^2} \) in terms of \( x \) and \( y \) is: \[ \frac{d^{2} y}{d x^{2}} = \frac{80xy^3 - \frac{75x^8}{4y^3}}{16y^6} \]