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Let $y = \frac{1}{\sqrt{1+2x}}$.

Prove: $(1+2x) y^{(n+1)}+ (2n+1) y^{(n)} = 0$

I can not understand that how it is proved, so please somebody help me.

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    Where did this $n$ come from?Is this the full question? – 2017-02-12
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    @takatiki,n is a positive integer (odd or even) – 2017-02-12
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    You need to prove that there exits such $n$? – 2017-02-12
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    For prove using Leibnitz's formula to find nth derivative – 2017-02-12
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    Are you sure you wrote the question correctly? – 2017-02-12

1 Answers 1

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The n th derivative of $y$ is $y^{(n)} = a_n (1+2x)^{-1/2 - n} $ with some constant $a_n$. Then $y^{(n+1)} = a_n \cdot (-1- 2n) \cdot (1+2x)^{1/2 - n-1} = \frac{-1- 2n}{1+2 x} y^{(n)} $. This proves the claim.

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    I need more answer. – 2017-02-13
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    You need to say what exactly you don't understand. – 2017-02-13
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    This prove by libneitz's formula of two function. – 2017-02-13