The Purplemath Forums

by **Andromeda** on Tue Nov 23, 2010 6:37 pm

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.

**Sponsor**

by **nona.m.nona** on Tue Nov 23, 2010 9:06 pm

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:

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

by **Andromeda** on Wed Nov 24, 2010 1:22 am

Ah I see. Thank you.

The Purplemath Forums

Implicit differentiation problem

Implicit differentiation problem

Re: Implicit differentiation problem

Re: Implicit differentiation problem