I need to show that in: x^3 + y^3 = 6xy the 2nd derivative is: Dn^2 = 16xy (2x - y^2) ^-3 how would i solve this?
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Originally Posted by mandn I need to show that in: x^3 + y^3 = 6xy the 2nd derivative is: Dn^2 = 16xy (2x - y^2) ^-3 how would i solve this? Differentiate implicity the first derivatiev would be $\displaystyle 3x^2+3y^2\frac{dy}{dx}=6x\frac{dy}{dx}+6y$ now solve for $\displaystyle \frac{dy}{dx}$ and then do the same again
ok lets try that
ok i got that part but what do u do after this step ?
Originally Posted by mandn ok i got that part but what do u do after this step ? After what step? if you are talking abou the step I showed you you should take the derivative again and solve for $\displaystyle \frac{dy^2}{d^2x}$ and remember you will get a couple $\displaystyle \frac{dy}{dx}$ you will have to sub these for your solved first derivative
ok thanks
Originally Posted by mandn ok thanks If you have any specific problems post them and we will do our best to walk you through them
right i will. but need to try solving it first.
k good luck
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