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Differential calculus

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    I'm stuck with a function of a function?

    I think I use the chain rule dy/dx = dy/du X du/dx

    But I don't know what to do. Could someone explain it to me really simply please?
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    Say you have y = sin(4x).

    A variation of the chain rule is to let u = 4x, v = sin(u).

    dy/dx = dv/dx = dv/du \times du/dx = cos(u) \times 4 = 4cos(u) = 4cos(4x)

    Does that make sense?
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    (Original post by jami74)
    I'm stuck with a function of a function?

    I think I use the chain rule dy/dx = dy/du X du/dx

    But I don't know what to do. Could someone explain it to me really simply please?
    Its better that you give the question you are stuck on.
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    (Original post by ViralRiver)
    Say you have y = sin(4x).

    A variation of the chain rule is to let u = 4x, v = sin(u).

    dy/dx = dv/dx = dv/du \times du/dx = cos(u) \times 4 = 4cos(u) = 4cos(4x)

    Does that make sense?
    I'm looking at it very hard. If I keep looking it might make sense and then I can apply it to my question.


    (Original post by raheem94)
    Its better that you give the question you are stuck on.
    Thank-you. It is y=cos^2x

    So is u = cos^2?
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    (Original post by jami74)
    I'm looking at it very hard. If I keep looking it might make sense and then I can apply it to my question.




    Thank-you. It is y=cos^2x

    So is u = cos^2?
     y = cos^2x  \\ u=cosx \\ y=u^2

    Now find  \displaystyle \frac{dy}{du} \ \ and \ \ \frac{du}{dx} .

    At the end, use the chain rule,  \displaystyle \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}
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    (Original post by raheem94)
     y = cos^2x  \\ u=cosx \\ y=u^2

    Now find  \displaystyle \frac{dy}{du} \ \ and \ \ \frac{du}{dx} .

    At the end, use the chain rule,  \displaystyle \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}
    Thank-you.

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Updated: May 11, 2012
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