SOLUTION: Hi, Totally Stuck here, not sure if the lecturer has skiped over this part, Find dy=dx as a function of x and y when: (a) y - x sin y - 3x^2 - 1 = 0 (b) x^2y^3 - x^2 + 3y -

Algebra ->  Algebra  -> Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: Hi, Totally Stuck here, not sure if the lecturer has skiped over this part, Find dy=dx as a function of x and y when: (a) y - x sin y - 3x^2 - 1 = 0 (b) x^2y^3 - x^2 + 3y -       Log On

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Question 375858: Hi, Totally Stuck here, not sure if the lecturer has skiped over this part,
Find dy=dx as a function of x and y when:
(a) y - x sin y - 3x^2 - 1 = 0
(b) x^2y^3 - x^2 + 3y - 3 = 0
(c) (1 - x^2) tan^-1 y = 3
Thanks
Answer by Fombitz(13828) About Me  (Show Source):
You can put this solution on YOUR website!
Use implicit differentiation, treat each terms as if it is differentiable (other than constants). Use the product rule, etc. to simplify. Gather up all of the dy and dx terms and then form the quotient, dy/dx.
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I'll do one, you do the rest.
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y+-+x%2Asin%28y%29+-+3x%5E2+-+1=0
dy-%28x%2Acos%28y%29dy%2Bsin%28y%29%2Adx%29-6xdx=0
dy-x%2Acos%28y%29dy=6xdx%2Bsin%28y%29dx
dy%281-x%2Acos%28y%29%29=%286x%2Bsin%28y%29%29dx
highlight%28dy%2Fdx=%286x%2Bsin%28y%29%29%2F%281-x%2Acos%28y%29%29%29