Respuesta :
[tex]\bf 4x^3+5y^3=4\implies 12x^2+15y^2\cfrac{dy}{dx}=0\implies 4x^2+5y^2\cfrac{dy}{dx}=0
\\\\\\
5y^2\cfrac{dy}{dx}=-4x^2\implies \boxed{\cfrac{dy}{dx}=\cfrac{-4x^2}{5y^2}}\\\\
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\cfrac{d^2y}{dx^2}=\stackrel{quotient~rule}{\cfrac{-8x(5y^2)~~-~~(-4x^2)(10y)\left( \frac{dy}{dx} \right)}{(5y^2)^2}}
\\\\\\
\cfrac{d^2y}{dx^2}=\cfrac{-40xy^2+(40x^2y)\left( \frac{-4x^2}{5y^2} \right)}{25y^4}[/tex]
[tex]\bf \cfrac{d^2y}{dx^2}=\cfrac{-40xy^2+(40x^2y)\left( \frac{-4x^2}{5y^2} \right)}{25y^4} \\\\\\ \cfrac{d^2y}{dx^2}=\cfrac{-40xy^2-\frac{160x^4y}{5y^2}}{25y^4}\implies \cfrac{d^2y}{dx^2}=\cfrac{\frac{-200xy^4-160x^4y}{5y^2}}{25y^4} \\\\\\ \cfrac{d^2y}{dx^2}=\cfrac{-200xy^4-160x^4y}{(25y^4)(5y^2)}\implies \cfrac{d^2y}{dx^2}=\cfrac{-200xy^4-160x^4y}{125y^6} \\\\\\ \cfrac{d^2y}{dx^2}=\cfrac{-40xy^4-32x^4y}{25y^6}[/tex]
[tex]\bf \cfrac{d^2y}{dx^2}=\cfrac{-40xy^2+(40x^2y)\left( \frac{-4x^2}{5y^2} \right)}{25y^4} \\\\\\ \cfrac{d^2y}{dx^2}=\cfrac{-40xy^2-\frac{160x^4y}{5y^2}}{25y^4}\implies \cfrac{d^2y}{dx^2}=\cfrac{\frac{-200xy^4-160x^4y}{5y^2}}{25y^4} \\\\\\ \cfrac{d^2y}{dx^2}=\cfrac{-200xy^4-160x^4y}{(25y^4)(5y^2)}\implies \cfrac{d^2y}{dx^2}=\cfrac{-200xy^4-160x^4y}{125y^6} \\\\\\ \cfrac{d^2y}{dx^2}=\cfrac{-40xy^4-32x^4y}{25y^6}[/tex]
