What is the answer

Question

Find the derivative d²y/dx² of the implicit curve y³ = xy + 1 at (0, 1)

Answer 1

We use implicit differentiation: 3y²y' = y + xy'. From here y^' = y/(3y² - x).

Now we have y'' = [y'(3y² - x) - (6yy' - 1)y]/(3y² - x)². Now for y' we substitude result of 1st step:

y'' = [2y - 6y³/(3y² - x)]/(3y² - x)² = - 2xy/(3y² - x)³.

At (0, 1) y'' = - 2·0·1/(3·1² - 0)³ = 0.

Answer 2

This is messy but not difficult.

3y² (dy/dx) - x(dy/dx) - y = 0

dy/dx = y/(3y² - x)

Now use the quotient rule to get d²y/dx²

You will get a fraction containing dy/dx.

You need to evaluate dy/dx at (0,1) and plug in the values to the messy fraction for the 2nd

derivative.

Quotient rule: downsie d upsie - upsie d downsie divided by downsie squared