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March 27, 2017

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find y' by implicit differentiation.

i don't understand, its so complicate.

1. 4cos(x-y)=3ysinx
2. 4sin(x-y) =3ysinx

  • Calculus implicit - ,

    4sin(x-y)*d(x-y)=3y' + 3ycosx

    On the left I applied the chain rule, and on the right i used the product rule.

    4sin(x-y)*(1-y')=3y' + 3ycosx

    Differentiate x - y and get 1 - y'

    Then do algebra to solve for y'

    4sin(x-y) - y'*4sin(x-y) = 3y' + 3ycosx

    3y' + y'*4sin(x-y) = 4sin(x-2) - 3ycosx

    y' = [4sin(x-2)-3ycosx]/[3+4sin(x-y)]

    Try the other one yourself.

  • Calculus implicit - ,

    ok i'll try another, thanks

  • Calculus implicit - ,

    oh
    d/dx cos u = - sin u du/dx
    You forgot the - sign

  • Calculus implicit - oops - ,

    also missed on the right side :-(

    4cos(x-y)=3ysinx
    -4sin(x-y)(1-y') = 3sinx*y' + 3ycosx
    -4sin(x-y) + 4sin(x-y)*y' = 3sinx*y' + 3ycosx
    y'(4sin(x-y) - 3sinx) = 3ycosx + 4sin(x-y)

    3ycosx + 4sin(x-y)
    ----------------------- = y'
    4sin(x-y) - 3sinx

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