How do you find $(d^2y)/(dx^2)$ for $x^3+4y^2=1$ ?

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Answer 1 · 2017-08-19T22:53:17

$(d^2y)/dx^2=-(12xy^2+9x^4)/(16y^3)$

Find $dy/dx$ first. Remember to use the chain rule when differentiating a function of $y$ .

$d/dx(x^3+4y^2)=d/dx(1)$

$3x^2+8ydy/dx=0$

$dy/dx=(-3x^2)/(8y)$

Now we need to differentiate again. In my opinion, the differentiation is simpler as such:

$dy/dx=-3/8x^2y^-1$

Now we need the product rule:

$(d^2y)/dx^2=-3/8(d/dxx^2)y^-1-3/8x^2(d/dxy^-1)$

$(d^2y)/dx^2=-3/8(2x)y^-1-3/2x^2(-y^-2)dy/dx$

Recall that $dy/dx=(-3x^2)/(8y)$ :

$(d^2y)/dx^2=(-3x)/(4y)+(3x^2)/(2y^2)((-3x^2)/(8y))$

$(d^2y)/dx^2=(-3x)/(4y)-(9x^4)/(16y^3)$

Getting a common denominator:

$(d^2y)/dx^2=-(12xy^2+9x^4)/(16y^3)$

How do you find (d^2y)/(dx^2)(d^2y)/(dx^2) for x^3+4y^2=1x^3+4y^2=1?