Question

Consider the curve defined implicitly by x^3 + y^3 = x^2 + y^2

a) Find dy/dx at the point(1,1).

Answer 1

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Answer 2

differentiating

3x^2 + 3y^2 * dy/dx = 2x + 2y*dy/dx

what we are doing is applying the chain rule for the terms in y, considering y to be a function of x

now make dy/dx the subject and plug in the given values of x and y

Answer 3

3x^2 + 3y^2 * 2x + 2y ?

Answer 4

3y^2 * dy/dx - 2y*dy/dx = 2x - 3x^2

dy /dx = 2x - 3x^2
----------
2y^2 - 2y

Answer 5

now plug in y=1 , x = 1

Answer 6

* sorry the denominator is 3y^2 - 2y not 2y^2 - 2y

Answer 7

but how you find this sir 3y^2 * dy/dx - 2y*dy/dx = 2x - 3x^2
I don't understand :(

Answer 8

go back to my first post
in order to get all the terms in dy.dx to the left side of the '='
i subtracted 3x^2 form the LHS and 2y*dy/dx from RHS

just straight forward algebra

Answer 9

sorry to be more accurate i should have said

subtract 3x^2 and 2y*dy/dx form both sides of the eqaution

Answer 10

thanks I understand sir :)

Answer 11

good (except I'm a ma'am) but no probs

Answer 12

haha sorry ma'ma