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Math Help - derivatives + tangent line equation

  1. #1
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    derivatives + tangent line equation

    For each of the following, compute \frac{dy}{dx} and find the equation for the tangent line at the indicated point x0. You may leave your answers in point-slope form

    1. y = x^2 + cos(pi(x^2+1)); x0 = 1

    for #1.

    I found \frac{dy}{dx} and used the chain rule for the cosine part and ended up with
    2x

    do i then use the tangent line formula ?
    y-y1 = m (x-x1)?

    if so, how do i plug in the numbers?
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  2. #2
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    you should have ended up with y' = 2x - sin(pi(x^2 + 1)) (2x * pi)

    so for x_0 = 1 you have the slope of he point (1,2) to be m=2 .... now you can find the pint slope form equation
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  3. #3
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    Quote Originally Posted by Arturo_026 View Post
    you should have ended up with y' = 2x - sin(pi(x^2 + 1)) (2x * pi)

    so for x_0 = 1 you have the slope of he point (1,2) to be m=2 .... now you can find the pint slope form equation
    am i not supposed to use the chain rule twice for the (pi(x^2+1)
    this is what i did,

    y' = 2x - sin(pi(x^2+1)) * (pi(x^2+1))' * (x^2+1)'
    y' = 2x - sin(pi(x^2+1)) * (0(x^2+1)) * 2x
    y' = 2x - sin(pi(x^2+1)) * 0 * 2x
    y' = 2x - 0
    y' = 2x
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  4. #4
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    Quote Originally Posted by break View Post
    am i not supposed to use the chain rule twice for the (pi(x^2+1)
    this is what i did,

    y' = 2x - sin(pi(x^2+1)) * (pi(x^2+1))' * (x^2+1)'
    y' = 2x - sin(pi(x^2+1)) * (0(x^2+1)) * 2x
    y' = 2x - sin(pi(x^2+1)) * 0 * 2x
    y' = 2x - 0
    y' = 2x
    Don't forget that pi is a number:

    for example if you are to find the derivative of 3(x^2 + 1) you don't use the chain rule like this: [3(x^2 + 1)]' * 2x , you see?

    you just take the 3 out of the differentiation and have: 3 * (x^2 + 1)' ; which gives you 6x
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  5. #5
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    Quote Originally Posted by Arturo_026 View Post
    Don't forget that pi is a number:

    for example if you are to find the derivative of 3(x^2 + 1) you don't use the chain rule like this: [3(x^2 + 1)]' * 2x , you see?

    you just take the 3 out of the differentiation and have: 3 * (x^2 + 1)' ; which gives you 6x
    oh okay, i know pi is a number, but dont i have to find the derivative of that too? since [pi(x^2+1)]' looks like product rule, then the derivative of pi = 0, and anything times zero is zero. why don't i find the derivative of pi?
    Last edited by break; December 6th 2009 at 11:24 AM.
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  6. #6
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    In any case, to avoid confusion why don't just multiply pi times (x + 1) and you'll see that (pi(x+1))' is the same as (x*pi)' + pi'
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