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calculus high school

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using implicit differentation, how do i solve this problem. it is a textbook example but the partwhere i "use the chain rule on the first term" doesn't make sense to me because they get

d/dx 7y^4+(x^3dy/dx +y d/dx x^3)+d/dx x

  • calculus high school -

    I get

    28y^3(dy/dx) + x^3(dy/dx) + y(3x^2) + 1 = 0
    dy/dx(28y^3 + x^3) = - 1 - 3(x^2)(y)

    dy/dx = (- 1 - 3(x^2)(y))/(28y^3 + x^3)

    I used he chain rule on the first term, the product rule on the second term, the others were routine

  • calculus high school -

    thanks. you got the answer right. but i don't understand how the chain rule is used iin this problem.

  • calculus high school -

    in 7y^4, according to "chain rule"
    exponent times front coefficient ---> 4(7) = 28
    keep the base, reduce the exponent by 1 ----y^4 becomes y^3
    times the derivative of the base, the base is y, so derivative of y is called dy/dx
    result ---> 28y^3(dy/dx)

  • calculus high school -

    isn't that the power rule?
    anyway, ow do you derive the second term?

  • calculus high school -

    never mind, the second term is derived using the product rule

  • calculus high school -

    Chain Rule with single variable:
    d/dx [3(2x-4)^2]

    Chain Rule with multiple variables:
    d/dx [3y^2]

    The y and the (x-4) terms serve a similar function. You technically use the chain rule all the time. The derivative of a variable, like x, is simply 1. So taking the derivative of 3x^2 is really easy.
    d/dx [3x^2]

    But (2x-4) and y aren't the same as x, and neither are the derivatives. So the chain rule will produce something appreciable. The derivative of 2x-4 is 2, so it affects the expression (whereas the derivative of x is simply one, and multiplying by one changes nothing). Y is also not equal to x. Because we are differentiating with respect to x, d/dx of y cant be 1. And we don't know anything else about y, so all we can do is say that dy/dx equals... well dy/dx.

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