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x3y2 + 2xy = 3.,
X=1,y = 2​


Sagot :

D ( x3 + y3 ) = D ( 4 ) ,

D ( x3 ) + D ( y3 ) = D ( 4 ) ,

(Remember to use the chain rule on D ( y3 ) .)

3x2 + 3y2 y' = 0 ,

so that (Now solve for y' .)

3y2 y' = - 3x2 ,

and

$ y' = \displaystyle{ - 3x^2 \over 3y^2 } = \displaystyle{ - x^2 \over y^2 } $