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Math Help - calculus please, higher order derivatives

  1. #1
    Newbie amanda1410's Avatar
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    calculus please, higher order derivatives

    Neem some urgent help on these please, if possible can you show me the method also so I can work out simular ones please

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  2. #2
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    Quote Originally Posted by amanda1410 View Post
    Neem some urgent help on these please, if possible can you show me the method also so I can work out simular ones please

    I will do,
    F'''(x)

    Do you see how to take derivative?
    You need derivative of,
    F''(x)=-2x(1+x^2)^{-2}
    Use product rule,
    (-2x)'(1+x^2)^{-2}+(-2x)[(1+x^2)^{-2}]'
    Thus, use chain rule on the second one,
    (-2)(1+x^2)^{-2}+(-2x)(2x)(-2)(1+x^2)^{-3}
    Thus,
    -2(1+x^2)^{-2}-8x^2(1+x^2)^{-3}
    Thus,
    -2(1+x^2)^{-2}[1+4x(1+x^2)^{-1}]
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  3. #3
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    Hello, Amanda!

    This one of the more unpleasant functions to differentiate.
    I suggest using the Quotient Rule.
    . . Those negative exponents can be dangerous.

    It starts simply enough . . .

    f(x)\;=\;\tan^{-1}(x)

    f'(x)\;=\;\frac{1}{x^2+1}\;=\;(x^2+1)^{-1}

    f''(x)\;=\;-1(x^2+1)^{-1}\cdot2x \;=\;-2\,\frac{x}{x^2+1}

    . . If there is a coefficient, I "pull it out front".

    Now it gets very messy . . . simplify each derivative before continuing.

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

    f'''(x)\;=\;-2\left[\frac{(x^2+1)\cdot1 - x\cdot2(x^2+1)\cdot2x}{(x^2+1)^4}\right]  \;=\;-2\left[\frac{(x^2+1)^2 - 4x^2(x^2+1)}{(x^2+1)^4}\right]

    Factor: . -2\,\frac{(x^2+1)\left(x^2+1 - 4x^2\right)}{(x^2+1)^4} \;= \;-2\,\frac{(1 - 3x^2)}{(x^2+1)^3} \;= \;2\,\frac{3x^2-1}{(x^2+1)^3}

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

    f^{(4)}(x) \;=\;2\,\left[\frac{(x^2+1)^3\cdot6x - (3x^2-1)\cdot3(x^2+1)^2\cdot2x}{(x^2+1)^6}\right]

    . . . . . =\;2\left[\frac{6x(x^2+1)^3 - 6x(3x^2-1)(x^2+1)^2}{(x^2+1)^6}\right]

    Factor: . 2\cdot6x(x^2+1)^2\left[\frac{(x^2+1) - (3x^2 - 1)}{(x^2+1)^6}\right] \;= \;12x\,\frac{2-2x^2}{(x^2+1)^4}\;= \;24\,\frac{x-x^3}{(x^2+1)^4}

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

    There were at least 10^{37} opportunities to make errors.
    . . I don't guarentee any of my results.

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  4. #4
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    Quote Originally Posted by Soroban View Post
    [size=3]Hello, Amanda!

    This one of the more unpleasant functions to differentiate.
    I suggest using the Quotient Rule.
    . . Those negative exponents can be dangerous.

    It starts simply enough . . .

    f(x)\;=\;\tan^{-1}(x)

    f'(x)\;=\;\frac{1}{x^2+1}\;=\;(x^2+1)^{-1}

    f''(x)\;=\;-1(x^2+1)^{-1}\cdot2x \;=\;-2\,\frac{x}{x^2+1}
    i think its -(x^2+1)^{-2}\cdot2x? is it?
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