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Differentiation watch

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    Could you give me a step-by-step on this differentiation:

    differentaite;
    1/(√(x+1))

    thank you.
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    Its just the chain rule.

     y = f(x)^n

     \frac{dy}{dx} = n(f(x)^{n-1}).f'(x)
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    (Original post by insparato)
    Its just

     y = f(x)^n

     \frac{dy}{dx} = n(f(x)^{n-1}).f'(x)
    But then its going to become way too complex;

    -0.5(x+1)^-0.75(0.5(x+1)-1.5)

    surely this is wrong???
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    Firstly if n = -0.5 what is (n-1) ? Apart from that the n(f(x)^{n-1}) term is correct.

    Secondly the f'(x) term I do not know where you get getting that from, if f(x) = x+1 then f'(x) is simply?
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    (Original post by TheLoneRanger)
    But then its going to become way too complex;

    -0.5(x+1)^-0.75(0.5(x+1)-1.5)

    surely this is wrong???
    Buh?

    You're differentiating (x+1)^{-1/2}. Take a deep breath and dive straight in - it's not as hard as it looks. (How would you differentiate (x+1)^3?)
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    yes basically using chain rule, use the power "-1/2" instead of the square root and hopefully you will see how to apply the chain rule.
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    (Original post by generalebriety)
    Buh?

    You're differentiating (x+1)^{-1/2}. Take a deep breath and dive straight in - it's not as hard as it looks. (How would you differentiate (x+1)^3?)
    Ok, overcomplicating it arent it , i just got it
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    (x+1)^-0.5

    let u = x+1

    dy/dx = -0.5(x+1)^-1.5
 
 
 
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