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)#