Question

Find y'' by implicit differentiation. 5x3 + 2y3 = 1

Answer 1

$\bf 5x^3+2y^3=1\implies 15x^2+\stackrel{chain-rule}{6y^2\cfrac{dy}{dx}}=0\implies 6y^2\cfrac{dy}{dx}=-15x^2 \\\\\\ \cfrac{dy}{dx}=\cfrac{-15x^2}{6y^2}\implies \cfrac{dy}{dx}=\cfrac{-5x^2}{2y^2}$

Answer 2

For clarity, please write x^3 and y^3, not x3 and y3.


5x^3 + 2y^3 = 1

Differentiating all terms with respect to x, 

15 x^2 + 6y^2(dy/dx) = 0

Rearranging, 6y^2(dy/dx) = -15x^2
                                                                 -15x^2
Dividing both sides by 6y^2,     dy/dx = ----------------- (which can be                                                                                     6y^2          reduced)

Now that you have dy/dx, differentiate again with respect to x.
Simplify your result.  This result represents   y".