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Math Help - inverse derivs...

  1. #1
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    inverse derivs...





    Consider the function below where x > 0.



    (a) Find and simplify f '(x).

    I need a little help with these bad johnnys...can someone show me how to do them, prefferably the steps...many thanks...

    mathaction
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  2. #2
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    Are you serious about needing help on g(x)? Take a good close look at it. You'll kick yourself.

    These are just Chain Rule exercises.

    For the first, you need to know these derivatives:

    \frac{d}{dt}\left(\frac{1}{t}\right)

    \frac{d}{dt}(arctan(t))

    \frac{d}{dt}(3t)

    Show us what you get.
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  3. #3
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    Quote Originally Posted by TKHunny View Post
    Are you serious about needing help on g(x)? Take a good close look at it. You'll kick yourself.

    These are just Chain Rule exercises.

    For the first, you need to know these derivatives:

    \frac{d}{dt}\left(\frac{1}{t}\right)

    \frac{d}{dt}(arctan(t))

    \frac{d}{dt}(3t)

    Show us what you get.
    yeah i already did the g(x), its just pi because sin and arcsin cancel out and the derivative of pi(x) is just pi cause pi is constant and d/dx is just 1

    d/dx(1/t) is 1/t^2, d/dx(arctan(t) is 1/(1+t^2), and d/dx(3t) is 3...now what do i do from here...
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  4. #4
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    Hello, mathaction!

    I must assume you know the derivative of the arctangent.

    . . \frac{d}{dx}(\arctan u) \;=\;\frac{u'}{1+u^2}


    f(x)\:=\:\frac{1}{\arctan(3t)}
    We have: . f(x)\;=\;[\arctan(3t)]^{-1}

    Chain Rule: . f'(x)\;=\;-1[\arctan(3t)]^{-2}\cdot\frac{1}{1+(3t)^2}\cdot 3 \;=\;\frac{-3}{[\arctan(3t)]^2(1 + 9t^2)}<br />



    g(x)\;=\;\sin\left[\arcsin(\pi x)\right]
    We have: . g(x)\;=\;\sin\left[\arcsin(\pi x)\right] \;=\;\pi x

    Therefore: . g'(x) \,=\,pi



    f(x) \;=\;\arctan(x) + \arctan\left(\frac{1}{x}\right)
    This one has a surprising answer . . .

    We have: . f'(x)\;=\;\frac{1}{1+x^2} + \frac{1}{1 + \left(\frac{1}{x}\right)^2}\cdot(-x^{-2}) \;= \;\frac{1}{1+x^2} + \frac{1}{1+\frac{1}{x^2}}\cdot\frac{-1}{x^2}

    . . = \;\frac{1}{1+x^2} - \frac{1}{x^2+1} \;=\;0


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    How come it works out to zero?

    Let \alpha = \arctan(x)
    Then: . \tan\alpha \:=\:\frac{x}{1}\:=\:\frac{opp}{adj}

    \alpha is in a right triangle with: opp = x,\;adj = 1
    Here is the triangle:
    Code:
                            *
                         *  *
                      *     *
                   *        * x
                *           *
             * α            *
          * * * * * * * * * *
                    1

    Let \beta = \arctan\left(\frac{1}{x}\right)
    Then: . \tan\beta \:=\: \frac{1}{x}\:=\:\frac{opp}{adj}
    \beta is in a right triangle with: opp = 1,\;adj = x
    Here is the triangle:
    Code:
                            *
                         * β*
                      *     *
                   *        * x
                *           *
             *              *
          * * * * * * * * * *
                    1


    Get it?
    \alpha and \beta are in the same right triangle.
    They are complementary: . \alpha + \beta \:=\:\frac{\pi}{2}

    We just differentiated a constant function: . f(x) \:=\:\frac{\pi}{2}

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