Hello

$\displaystyle sin(2x^3 + y^2) = cos(x^2 + 2y^3)$

How would this be solved using implicit differentiation?

Thank you

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- Nov 3rd 2008, 01:57 PMalgorithmImplicit differentiation
Hello

$\displaystyle sin(2x^3 + y^2) = cos(x^2 + 2y^3)$

How would this be solved using implicit differentiation?

Thank you - Nov 3rd 2008, 03:12 PMJameson
I'll do the left hand side and let you do the right.

$\displaystyle \cos(2x^3+y^2) (6x^2+2y*y') = ... $

Use the chain rule. The outermost function in this case was the sine function, then multiply by the derivative of the inner terms. Same thing for the right hand side.

Then, you'll have to do some algebra to solve for y'. - Nov 3rd 2008, 04:15 PMalgorithm
I see. How would you express the derivative of $\displaystyle cos(u)$ using the chain rule when $\displaystyle u = 2x^3 + y^2$?

Thank you - Nov 3rd 2008, 04:19 PMelpermic
$\displaystyle u' = 6x^2 + 2y'y$

$\displaystyle -sin(2x^3 + y^2) (6x^2 + 2y'y)$

Should be like that - Nov 3rd 2008, 04:51 PMalgorithm
Thank you. How would this application of the chain rule be written in the form: $\displaystyle dy/dx = du/dx * ...$, for example?