Question

How do you use implicit differentiation to find `(d^2y)/dx^2` of `x^3+y^3=1` ?

Answer 1

By implicitly differentiating twice, we can find
`{d^2y}/{dx^2}=-{2x}/y^5`.

First, let us find `{dy}/{dx}`.
`x^3+y^3=1`
by differentiating with respect to `x`,
`Rightarrow 3x^2+3y^2{dy}/{dx}=0`
by subtracting `3x^2`,
`Rightarrow3y^2{dy}/{dx}=-3x^2`
by dividing by `3y^2`,
`Rightarrow {dy}/{dx}=-{x^2}/{y^2}`

Now, let us find `{d^2y}/{dx^2}`.
by differentiating with respect to `x`,
#Rightarrow{d^2y}/{dx^2}=-{2x cdot y^2-x^2 cdot 2y{dy}/{dx}}/{(y^2)^2} =-{2x(y^2-xy{dy}/{dx})}/{y^4}#
by plugging in `{dy}/{dx}=-{x^2}/{y^2}`,
`Rightarrow{d^2y}/{dx^2}=-{2x[y^2-xy(-x^2/y^2)]}/y^4=-{2x(y^2+x^3/y)}/y^4`
by multiplying the numerator and the denominator by `y`,
`Rightarrow{d^2y}/{dx^2}=-{2x(y^3+x^3)}/y^5`
by plugging in `y^3+x^3=1`,
`Rightarrow{d^2y}/{dx^2}=-{2x}/y^5`