Find the second derivative d2y/dx2 of the function defined implicitly by the equation.

Question

Find the second derivative d2y/dx2 of the function defined implicitly by the equation.

x^1/3 + y^1/3 = 9

Answer 1

Apply the derivative operator:

d/dx(x^1/3 + y^1/3) = d/dx(9)

The derivative of x^1/3 (with respect to x) is 1/3x^-2/3

The derivative of y^1/3 (with respect to x) is 1/3y^-2/3•dy/dx

The derivative of 9 (with respect to x) is 0

Resulting in:

1/3x^-2/3 + 1/3y^-2/3•dy/dx = 0 [Now, solve for dy/dx - which is the first derivative:

1/3y^-2/3•dy/dx = - 1/3x^-2/3

dy/dx = - 1/3x^-2/3/1/3y^-2/3

dy/dx = - x^-2/3/y^-2/3

dy/dx = - y^2/3/x^2/3

To find the second derivative, again apply the derivative operator:

d²y/dx² = d/dx(- y^2/3/x^2/3)

The derivative of dy/dx with respect to x is d²y/dx² (the second derivative)

For the derivative of -y^2/3/x^2/3 we must use the quotient rule (f'g - g'f)/g² with as follows:

f is - y^2/3

f ' = -2/3y^-1/3•dy/dx

g is x^2/3

g ' is 2/3x^-1/3

g² is x^4/3

So the derivative becomes:

[-2/3y^-1/3•dy/dx•x^2/3 - 2/3x^-1/3 •(- y^2/3)]/x^4/3

So:

d²y/dx² = -2/3[dy/dx•x^2/3/y^1/3 - y^2/3]/x^5/3

But dy/dx = - y^2/3/x^2/3 so subbing that in we get:

At this point it gets confusing unless you switch to a more orderly numerator and denominator that are written over each other so I'll switch:

Then getting a common denominator on the top results in:

The original equation tells us x^1/3 + y^1/3 = 9 so subbing that in, we get:

d²y/dx² = 6y^1/3/x^5/3