Question

Find y'' by implicit differentiation. 4x^3 + 5y^3 = 4

Answer 1

`\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}}\\\\ -------------------------------\\\\ \cfrac{d^2y}{dx^2}=\stackrel{quotient~rule}{\cfrac{-8x(5y^2)~~-~~(-4x^2)(10y)\left( (dy)/(dx) \right)}{(5y^2)^2}} \\\\\\ \cfrac{d^2y}{dx^2}=\cfrac{-40xy^2+(40x^2y)\left( (-4x^2)/(5y^2) \right)}{25y^4}`


`\bf \cfrac{d^2y}{dx^2}=\cfrac{-40xy^2+(40x^2y)\left( (-4x^2)/(5y^2) \right)}{25y^4} \\\\\\ \cfrac{d^2y}{dx^2}=\cfrac{-40xy^2-(160x^4y)/(5y^2)}{25y^4}\implies \cfrac{d^2y}{dx^2}=\cfrac{(-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}`