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I found out that derivative of a function at its Root with degree more than 1 Is Zero.

For example :

at x = 1 : $$[log(x)]^n , n > 1$$ at x = $Pi/2$ $$[cos(x)]^n , n > 1$$ at x = 1 $$log(x).(e^x - e) $$ at x = 0 $$log(x+1).(e^x-1)$$ Or any polynomial

But I can't find a proof for it and I would appreciate it if anyone has an example that contradicts it.

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    Are you saying that if $f(\alpha)=0$, then $\frac{df^n}{dx}(\alpha)=0$? Hint: use the chain rule. – 2017-01-24
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    @MichaelBurr I'm not saying that the function might be like log(x)*(e^x - e) at 0 – 2017-01-24
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    All of your cases can be analyzed with the product rule. Whenever you apply the product rule, you will be left with a factor in the original product, which vanishes. – 2017-01-24
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    :-) This is how one usually defines multiplicity of roots in non-polynomial functions. – 2017-01-24

2 Answers 2

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Stated more precisely, suppose $f(x)=A(x)B(x)C(x)$, where $A(x_0)=0$ and $B(x_0)=0$ and $A$, $B$ and $C$ are differentiable at $x_0$ ($C$ could be constant).

Then $$ f'(x_0)=A'(x_0)B(x_0)C(x_0)+A(x_0)B'(x_0)+A(x_0)B(x_0)C'(x_0)=0 $$

In your first case, $(\log x)^n$, for $n\ge2$, take $A(x)=\log x$, $B(x)=\log x$ and $C(x)=(\log x)^{n-2}$.

In the case of $(\log x)(e^x-e)$, take $A(x)=\log x$, $B(x)=e^x-e$ and $C(x)=1$.

You can't apply this to $(\sin\sqrt{x})^2$ at $0$, because $\sin\sqrt{x}$ is not differentiable at $0$.

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There is a subtle exception to your claim. Given a function $g(x) = [f(x)]^n, n \gt 1$, your claim is that $g'(\alpha)=0 ,\forall \alpha,where f(\alpha)=0$. However,note that:

$g'(\alpha)= n.[f(\alpha)]^{n-1}.f'(\alpha)$

The rhs,may not exist if $f'(\alpha)$ is infinite(indeterminate form). Given $f'(\alpha)$ exists and is finite,your claim is true, but not otherwise.

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    Can you give an example please. – 2017-01-24
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    An example is the only thing not coming to my mind right now. Ill post as soon i find one. – 2017-01-24
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    Consider the function $g(x) = sin^2(\sqrt x)$ x = 0 is a root. But $g'(x) = sin(\sqrt x).cos(\sqrt x)/(\sqrt x)$ note that at x= 0 this is indeterminate. Even the limiting value is 1 and not zero. – 2017-01-24