Question

How do you find the second derivative by implicit differentiation on `x^3y^3=8` ?

Answer 1

As the first step, we will differentiate once, and apply the product rule:

`d/dx[x^3]*y^3 + d/dx[y^3]*x^3 = d/dx[8]`

For `y^3`, remember to use the chain rule. Simplifying yields:

`3x^2y^3 + 3y^2x^3dy/dx = 0`

Now, we will solve for `dy/dx`:

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

We can cancel off the 3, an `x^2`, and a `y^2`, which will yield:

`dy/dx = -y/x`

Now, differentiate once again. We will apply the quotient rule:

`(d^2y)/(dx^2) = -(x*dy/dx - y*1)/x^2`

Looking back at the previous equation for `dy/dx`, we can substitute into our equation for the second derivative to get it in terms of only `x` and `y`:

`(d^2y)/(dx^2) = -(x*(-y/x) - y*1)/x^2`

Simplifying yields:

`(d^2y)/(dx^2) = (2y)/x^2`