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Math Help - [SOLVED] Differentiate √(x + 1) with respect to x.

  1. #1
    Member looi76's Avatar
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    [SOLVED] Differentiate √(x + 1) with respect to x.

    Question:
    Differentiate \sqrt{(x^2 + 1)} with respect to x.

    Attempt:

    \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}

    u = \sqrt{(x^2 + 1)} = (x^2 + 1)^{\frac{1}{2}}

    \frac{dy}{du} = x^2

    \frac{du}{dx} = \frac{1}{2}u^{-\frac{1}{2}}

    \frac{dy}{dx} = x^2 \times \frac{1}{2}u^{-\frac{1}{2}}

    \frac{dy}{dx} = x^2 \times \frac{1}{2}(x^2 + 1)^{-\frac{1}{2}}

    Are my steps right? and can you please complete the steps? thnx!
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  2. #2
    Member Danshader's Avatar
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    for the final answer it is 2x instead of x^2:

    dy/dy = 2x(1/2)(x^2+1)^(-1/2)
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  3. #3
    Member looi76's Avatar
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    Quote Originally Posted by Danshader View Post
    for the final answer it is 2x instead of x^2:

    dy/dy = 2x(1/2)(x^2+1)^(-1/2)
    Why is it 2x instead of x^2 ?
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  4. #4
    Moo
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    Hello,

    Because the derivative for x+1 is 2x
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  5. #5
    Member Danshader's Avatar
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    take for example X^n

    differentiating it w.r.t x will give:
    n(d/dx X)(X)^(n-1)



    take for example (X+1)^n

    differentiating it w.r.t x will give:
    n(d/dx (X+1))(X+1)^(n-1)



    take for example ((X^m)+1)^n

    differentiating it w.r.t x will give:
    n(d/dx ((X^m)+1))((X^m)+1)^(n-1)

    in youe question n = 1/2, m=2.
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  6. #6
    Member disclaimer's Avatar
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    Quote Originally Posted by looi76 View Post
    Question:
    Differentiate \sqrt{(x^2 + 1)} with respect to x.

    Attempt:

    \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}

    u = \sqrt{(x^2 + 1)} = (x^2 + 1)^{\frac{1}{2}}

    \frac{dy}{du} = x^2

    \frac{du}{dx} = \frac{1}{2}u^{-\frac{1}{2}}

    \frac{dy}{dx} = x^2 \times \frac{1}{2}u^{-\frac{1}{2}}

    \frac{dy}{dx} = x^2 \times \frac{1}{2}(x^2 + 1)^{-\frac{1}{2}}

    Are my steps right? and can you please complete the steps? thnx!
    What a strange notation you have there...
    I would write immediately:
    \left(\sqrt{(x^2 + 1)}\right)' = \left((x^2 + 1)^{\frac{1}{2}}\right)'=\frac{1}{2}(x^2 + 1)^{-\frac{1}{2}}\cdot{2x}=x (x^2 + 1)^{-\frac{1}{2}}=\frac{x}{\sqrt{x^2 + 1}}
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