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Math Help - derivative check

  1. #1
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    derivative check

    f(x) = (sqrt(x))/(x^3+1)
    f'(x) = (-2.5x^5/2)+(1/2sqrt(x)) / (x^3+1)^2

    f(x) = (cosx/t^3)
    f'(x) = (-x^3sinx)-(3xcosx)/(t^3)

    f(x) = [2-(1/x)]/(x-3)
    f'(x) = (-2x+2x-3)/(x-3x)


    unsuree if i have done these correctly on my H/W...-thx
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  2. #2
    Member mathemagister's Avatar
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    Quote Originally Posted by maybnxtseasn View Post
    f(x) = (sqrt(x))/(x^3+1)
    f'(x) = (-2.5x^5/2)+(1/2sqrt(x)) / (x^3+1)^2

    f(x) = (cosx/t^3)
    f'(x) = (-x^3sinx)-(3xcosx)/(t^3)

    f(x) = [2-(1/x)]/(x-3)
    f'(x) = (-2x+2x-3)/(x-3x)


    unsuree if i have done these correctly on my H/W...-thx
    Here's a kickstart on the first one:

    f(x) = \frac{\sqrt x}{x^3+1}

    f'(x) = \frac {(x^3+1) \frac{1}{2}x^{-\frac{1}{2}} - \sqrt{x} \cdot 3x^2}{(x^3+1)^2} = \frac {\frac{1}{2}x^{\frac{5}{2}} + \frac{1}{2}x^{-\frac{1}{2}} - 3x^{\frac{5}{2}}}{(x^3+1)^2} = \frac {-\frac{5}{2}x^{\frac{5}{2}} + \frac{1}{2}x^{-\frac{1}{2}}}{(x^3+1)^2}

    I can't read your Math clearly, but I think you got it right. By
    (1/2sqrt(x))
    , you mean \frac{1}{2\sqrt{x}} right? Then, I think what you have is correct.

    Hope that was helpful
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  3. #3
    Member mathemagister's Avatar
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    Quote Originally Posted by maybnxtseasn View Post
    f(x) = (sqrt(x))/(x^3+1)
    f'(x) = (-2.5x^5/2)+(1/2sqrt(x)) / (x^3+1)^2

    f(x) = (cosx/t^3)
    f'(x) = (-x^3sinx)-(3xcosx)/(t^3)

    f(x) = [2-(1/x)]/(x-3)
    f'(x) = (-2x+2x-3)/(x-3x)


    unsuree if i have done these correctly on my H/W...-thx
    And for the second one:
    f(x) = \frac{cosx}{t^3} = t^{-3}cosx

    You are forgetting implicit differentiation here. You are differentiating with respect to x, so when you take the derivative of t it will be t' or \frac {dt}{dx}, but not 1.

    f'(x) = -t^{-3}sinx - 3t^{-4}cosx \cdot t'

    You also seem to be changing the t into an x sometimes. Make sure you understand what I did there. If you want me to explain it in more detail, I will.
    Hope I helped you
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  4. #4
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    Hello, maybnxtseasn!

    I think I agree with your first answer . . .


    f(x) \:= \:\frac{x^{\frac{1}{2}}} {x^3+1}

    f'(x) \:=\: \frac{-2.5x^{\frac{5}{2}} +{\color{red}\dfrac{1}{2\sqrt{x}}}}{(x^3+1)^2} . . Is this what you meant?

    f'(x) \;=\;\frac{(x^3+1)\cdot \frac{1}{2}x^{-\frac{1}{2}} - x^{\frac{1}{2}}(3x^2)}{(x^3+1)^2} \;=\;\frac{\frac{1}{2}x^{-\frac{1}{2}}(x^3+1) - 3x^{\frac{5}{2}}}{(x^3+1)^2}


    Multiply by \frac{2x^{\frac{1}{2}}}{2x^{\frac{1}{2}}}

    . f'(x) \;=\;\frac{2x^{\frac{1}{2}} \bigg[ \frac{1}{2}x^{-\frac{1}{2}}(x^3+1) - 3x^{\frac{5}{2}} \bigg]}  {2x^{\frac{1}{2}}\cdot (x^3+1)^2}

    . . . . =\;\frac{x^3+1 - 6x^3}{2x^{\frac{1}{2}}(x^3+1)^2} \;=\;\frac{-5x^3 + 1}{2\sqrt{x}\,(x^3+1)^2}

    . . which is equivalent to your result.

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  5. #5
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    yo

    Quote Originally Posted by mathemagister View Post
    And for the second one:
    f(x) = \frac{cosx}{t^3} = t^{-3}cosx

    You are forgetting implicit differentiation here. You are differentiating with respect to x, so when you take the derivative of t it will be t' or \frac {dt}{dx}, but not 1.

    f'(x) = -t^{-3}sinx - 3t^{-4}cosx \cdot t'

    You also seem to be changing the t into an x sometimes. Make sure you understand what I did there. If you want me to explain it in more detail, I will.
    Hope I helped you
    i still dont see how u got that awnser....anyone else help? how come i dont get this awnser when i do the simple quotient rule with this?
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