Question

If x^3+y^3=8 show that the second derivative of y with respect to x is -16/y^5

Answer 1

This is implicit differentiation but let me solve this on paper first before I help

Answer 2

this involves solving for 1st derivative first.

differentiate a 2nd time, solve for 2nd derivative and substitute in what you get for 1st derivative

Answer 3

I am getting there on paper, just takes time to simplify

Answer 4

ok i'll let you show solution :)

Answer 5

For the first derivative I'm gettting dy/dx=3x^2/3y
Am i on the right track?

Answer 6

should be dy/dx = 3x^2/3y^2

Answer 7

by yes that is the right track

Answer 8

As far as i can tell from my work, the 2nd derivative will not equal -16/y^5

Answer 9

its a typo... should be -16x/y^5

Answer 10

Ok I figured it out hopefully, so as you did you found dy/dx (Still did not get that lol)

\[x^3 + y^3 =\]
\[3x^2 = 3y^2\frac{ dy }{ dx }= 0\]
\[\frac{ dy }{ dx } = \frac{ -3x^2 }{ 3y^2 }\]

Right?

Answer 11

woops suppose to equal 8 at top there

Answer 12

Which i assume you had just mistyped it...NOW comes the hard part the 2nd derivative. I rewrote it to do a product with chain rule although you can do a quotient rule

\[\frac{ dy }{ dx } = (-3x^2)(3y^2)^{-1}\]
\[\frac{ d^2y }{ dx^2 } = (-6x)(3y^2)^{-1} - (3y^2)^{-2}6y \frac{ dy }{ dx }-3x^2\]

You understand how I got to that?

Answer 13

you could also just cancel out the 3's

Answer 14

Yea but i realized that just now and i did my work without that so i am just following my work :P

Answer 15

@step8410 do you understand so far?