## Implicit differentiation problem

Limits, differentiation, related rates, integration, trig integrals, etc.

### Implicit differentiation problem

Find y" by implicit differentiation

x^3 + y^3 = 1

Here's my work:
3x^2 + 3y^2y' = 0

y' = (-3x^2)/(3y^2) = (-x^2)/(y^2)

y" = (-2xy^2 + 2x^2y(dy/dx)) / y^4

= 2(-xy - x^4y^-2) / y^3

The back of the book says that the answer is -2x / y^5. I don't understand how they're simplifying it to that. Thanks for any help.
Andromeda

Posts: 11
Joined: Tue Oct 12, 2010 1:56 am

### Re: Implicit differentiation problem

They appear to have used a couple substitutions and then simplified. Your answer is equivalent.

Andromeda wrote:y" = (-2xy^2 + 2x^2y(dy/dx)) / y^4

This is correct. However, try plugging the fractional form for dy/dx into the numerator, and simplifying:

. . . . .$-2xy^2\, +\, 2x^2 y\left(\frac{dy}{dx}\right)$

. . . . .$-2xy^2\, +\, 2x^2 y\left(\frac{-x^2}{y^2}\right)$

. . . . .$-2xy^2\, -\, \frac{2x^4}{y}$

. . . . .$\frac{-2xy^3}{y}\, -\, \frac{2x^4}{y}$

. . . . .$\frac{-2xy^3\, -\, 2x^4}{y}$

Plug this back into the expression for y":

. . . . .$\frac{\left(\frac{-2xy^3\, -\, 2x^4}{y}\right)}{y^4}$

. . . . .$\frac{-2xy^3\, -\, 2x^4}{y^5}$

Now factor:

. . . . .$\frac{-2x\left(y^3\, +\, x^3\right)}{y^5}$

Now substitute:

. . . . .$\frac{-2x(1)}{y^5}\, =\, \frac{-2x}{y^5}$
nona.m.nona

Posts: 256
Joined: Sun Dec 14, 2008 11:07 pm

### Re: Implicit differentiation problem

Ah I see. Thank you.
Andromeda

Posts: 11
Joined: Tue Oct 12, 2010 1:56 am