Implicit differentiation calculus

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PLEASE HELP , I DON'T UNDERSTAND

find dy/dx by implicit differentiation

a) xy + x^5y^2 + 3x^3 - 4 = 1

b) lny = cos x

thank you

  • Implicit differentiation calculus -

    normally you just have an equation like

    y = 3x^2
    and you take the derivative of both sides, to get
    y' = 6x

    Then you advance to using the chain rule and product rule, and you have something like

    y = e^x*sin^2(6x)
    and you take the derivative of both sides to get
    y' = e^x(sin^2(6x)+12sin(6x)cos(6x))

    implicit differentiation just uses these techniques. Take the derivative of both sides of the equation, using the chain rule, and remembering that dx/dx = x' = 1 and dy/dx = y'

    So, for your problems,

    a) the derivative of xy is

    d/dx(xy) = x'y + xy' = y + xy'
    d/dx(x^5y^2) = (5x^4)(y^2) + (x^5)(2yy')
    d/dx(3x^3) = 9x^2
    d/dx(-4) = 0
    d/dx(1) = 0
    and put it all together to get
    y + xy' + 5x^4y^2 + 2x^5yy' + 9x^2 = 0
    Now just solve for y' by collecting terms
    y'(x+2x^5y) + (y + 5x^4y^2 + 9x^2) = 0

    y' = -(y + 5x^4y^2 + 9x^2)/(x+2x^5y)

    b) again, use the chain rule:
    d/dx(ln y) = d/dy(ln y)*dy/dx = 1/y y'
    1/y y' = -sin x
    y' = -y sin x

  • Implicit differentiation calculus -

    thank for you so much steve!!!

    i just wanted to double check.. for (a) it is y' = - (y+5x^4y^2+9x^2)/(x+2x^5y)

    i was wondering if the last part of it (x+2x^5y) is it 2x^(5y) or 2x^5(y). im not sure if the "y" is in the exponent or not.

  • Implicit differentiation calculus -

    No, since we took the derivative of x^5*y^2 we get (5x^4)(y^2) + (x^5)(2yy')

    That is, 2(x^5)(y)

  • Implicit differentiation calculus -

    ok so the final answer would b

    a) y' = -(y + 5x^4y^2 + 9x^2)/(x+2x^5y)

    and

    b) y' = -y sin x

    thank you SO much once again steve, my exam is in a couple of hours and this solution really helped me, im very thankful. thanks so much!!

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