Implicit Differentiation, Find the second derivative.
xy+y^3=1
I found that the derivative of this is -y/x+3y^2
And then to find the second derivative I used the quotient rule.
(x+3y^2) ( -dy/dx) - (-y)(1+6y dy/dx) all over (x+3y)^2

I multiply those all together and I get
-x dy/dx - 3y^2dy/dx + y +6y^2 dy/dx
all over the original denominator
Simplifying it I got (-x-3y^2+6y^2)(dy/dx) +y
and then I get -x dy/dx + 3y^2 dy/dx+ y
multiplying it by the original dy/dx
xy/x+3y^2 - 3y^3/x+3y^2
I get xy-3y^3 / (x+3y)^3.

Question

Implicit Differentiation, Find the second derivative.
xy+y^3=1
I found that the derivative of this is -y/x+3y^2
And then to find the second derivative I used the quotient rule.
(x+3y^2) ( -dy/dx) - (-y)(1+6y dy/dx) all over (x+3y)^2

I multiply those all together and I get
 -x dy/dx - 3y^2dy/dx + y +6y^2 dy/dx
all over the original denominator
Simplifying it I got (-x-3y^2+6y^2)(dy/dx) +y
and then I get -x dy/dx + 3y^2 dy/dx+ y
multiplying it by the original dy/dx
xy/x+3y^2 - 3y^3/x+3y^2
I get xy-3y^3 / (x+3y)^3. <- Simplified denominators.
Need confirmation.

EDIT: Using the Quotient rule I ended up with 2x / (x+3y^2) ^3 ? correct no?

marsxtc · Answer

I usually think that since the original equation is xy+y^3 = 1, that for the very last equation on the numerator it should've been xy +y^3, but I'm unsure where I went wrong.

primeralph · Answer

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