1. Ha#TH y' H y'' 1.26. x^3+y^3=5x

To find the first and second derivatives of the given equation \(x^3 + y^3 = 5x\), we need to implicitly differentiate with respect to \(x\).<br /><br />First, let's differentiate both sides of the equation with respect to \(x\):<br /><br />\[ \frac{d}{dx}(x^3 + y^3) = \frac{d}{dx}(5x) \]<br /><br />Using the chain rule on the left side, we get:<br /><br />\[ 3x^2 + 3y^2 \frac{dy}{dx} = 5 \]<br /><br />Now, we can solve for \(\frac{dy}{dx}\):<br /><br />\[ 3y^2 \frac{dy}{dx} = 5 - 3x^2 \]<br /><br />\[ \frac{dy}{dx} = \frac{5 - 3x^2}{3y^2} \]<br /><br />So, the first derivative \(\frac{dy}{dx}\) is \(\frac{5 - 3x^2}{3y^2}\).<br /><br />Next, let's find the second derivative \(\frac{d^2y}{dx^2}\). We need to differentiate \(\frac{dy}{dx}\) with respect to \(x\):<br /><br />\[ \frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{5 - 3x^2}{3y^2} \right) \]<br /><br />Using the quotient rule for differentiation, we get:<br /><br />\[ \frac{d^2y}{dx^2} = \frac{(3y^2)(-6x) - (5 - 3x^2)(6y)}{(3y^2)^2} \]<br /><br />Simplifying the expression, we have:<br /><br />\[ \frac{d^2y}{dx^2} = \frac{-18xy^2 - 30y + 18x^2y}{9y^4} \]<br /><br />So, the second derivative \(\frac{d^2y}{dx^2}\) is \(\frac{-18xy^2 - 30y + 18x^2y}{9y^4}\).<br /><br />In summary, the first derivative \(\frac{dy}{dx}\) is \(\frac{5 - 3x^2}{3y^2}\) and the second derivative \(\frac{d^2y}{dx^2}\) is \(\frac{-18xy^2 - 30y + 18x^2y}{9y^4}\).