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Let $f(x)$ be a function such that it is $n$ times differentiable and $f^{'}(a)=f^{''}=(a)f^{'''}=(a)....=f^{n-1}(a)=0$ and $f^{n}\ne0.$

The $n^{th}$ derivative test tells us about the concavity of the derivative whether it has a local extremum or an inflection point at some $x=a$ depending on whether $n$ is even or odd.But can we predict the nature of $f^{n-1}(x)$ from this whether it is decreasing or increasing in the interval $x>a$ or $x<a$ or has a local extremum at $x=a$?

If $f^{n}(x)>0$, does it imply $f^{n-1}(x)$ is increasing?Is there a corelation between the $n^{th}$ derivatives of a function?

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Let $g(x)=f^{(n-1)}(x)$.

If $g'(x)>0$ (i.e. $f(^{(n)}(x)>0$) then obviously $g(x)$ (i.e. $f^{(n-1)}(x)$) is increasing.

The correlation between the derivatives of a function is... that they are derivatives of each other. But the values of the derivatives at a single point are independent and can be quite arbitrary. For example, the derivatives of a polynomial at $0$ are simply its coefficients (times a factorial).

You will understand the meaning of the derivative test by considering the monomial functions $f(x)=x^m$ for increasing $m$: minima and inflection points alternate.

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The way you ask the question doesn't quite make sense; $f^{(n)}(a)$ is one single number, it can't be increasing.

But the correlation you're asking about is just the fact that $f^{(n+1)}$ is the derivative of $f^{(n)}$. So for example if $f^{(n+1)}>0$ on an interval then $f^{(n)}$ is increasing on that interval, yes.

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Since $f^{n-1}$ is differentiable, therefore $f^{n-1}$ is continuous at $a$. Now you can assume that $f^{n-1}$ is continuous in some neighborhood of $a$ (otherwise there is not a lot of analysis we can do). Then since $f^{n-1}(a)=0$ it has a extrema at $a$ and so you do the same analysis as you would considering $f^{n-1}$ as a function $F$ (apply higher derivative tests.)

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