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The $ nth $ derivative of a function tells the rate of change of the $ (n-1)th $ derivative of a function, and if $n=1$, we'll get the first derivative of our function.

So consider a function
$y = x^2$
$\implies$ $\frac {d(x^2)}{dx} = 2x$
This means that the slope of the tangent line at any $x$ value of the graph is $2x$.

But what happens when we further differentiate our first derivative?
We have $\frac{d(2x)}{dx} = 2 $

So if we plot $y'(x)$ on the graph, we get to know that the slope of $y'(x)$ at any $x$ value is 2. This means that the graph of $y'(x)$ is a straight line. So my question is :

Is it true that every differentiable function's last derivative expression, when plotted, will be a straight line?

Or in other words, can every differentiable function be differentiated till we get a constant?

EDIT : Every differentiable function can be differentiated infinitely many times, and the derivative of $0$ is $0$. I thought that $0$ couldn't be differentiated further.

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    $\begingroup$ What do you mean by "last derivative"? There is no "last derivative" for $x^2$, it is differentiable infinitely many times. $\endgroup$
    – user223391
    Feb 26, 2017 at 15:21
  • $\begingroup$ @Zachary Selk. After differentiating 2, we get 0. How can we further differentiate 0? $\endgroup$
    – Saksham
    Feb 26, 2017 at 15:22
  • $\begingroup$ The derivative of $0$ is just $0$ $\endgroup$
    – user223391
    Feb 26, 2017 at 15:23
  • $\begingroup$ Ok thanks a lot for the help. So this means that after we have reached 0, we will continue to get 0 thereafter? $\endgroup$
    – Saksham
    Feb 26, 2017 at 15:25
  • $\begingroup$ Yes, that is true $\endgroup$
    – user223391
    Feb 26, 2017 at 15:25

1 Answer 1

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Consider $f(x) = e^x$. Every derivative of $f$ is also $e^x$, so it will never become constant. In fact, almost all functions have the property that no amount of differentiating will produce a constant.

It is true, however, that for any polynomial $g(x)$ of degree $n$, the $n$th derivative of $g$ is constant (and so the $n+1$-st derivative is the zero function).

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