Question

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

Answer 1

We first have to find the first derivative before we can find the second.  First derivative with respect to x is $9x^2+12y^2 \frac{dy}{dx}=0$ so $12y^2 \frac{dy}{dx}=-9x^2$  and  $\frac{dy}{dx}=- \frac{9x^2}{12y^2}=- \frac{3x^2}{4y^2}$.  That's the first derivative.  Now for the second it's probably easiest to use the quotient rule, even though it's usually long and drawn out.  That looks like this: $\frac{dy}{dx}= \frac{4y^2(-6x)-[-3x^2(8y \frac{dy}{dx})] }{(4y^2)^2}$  which simplifies a bit to  $\frac{dy}{dx}= \frac{-24xy^2+24x^2y( \frac{dy}{dx}) }{16y^4}$.  We will multiply both sides by that denominator to get rid of it which leaves us with  $16y^4 \frac{dy}{dx}=-24xy^2+24x^2y \frac{dy}{dx}$.  Get both dy/dx terms on the same side, and then factor it out.  $\frac{dy}{dx}(16y^4-24x^2y)=-24xy^2$.  Divide to isolate the dy/dx: $\frac{d^2y}{dx^2}= \frac{-24xy^2}{16y^4-24x^2y}$.  There's no simplifying you could do after that that would make any significant difference.