0
$\begingroup$

I am ask to find the most general antiderivative of $f(x)= x^n$ where $n \geq 0$.

However, I wondering how the derivative of $\dfrac{x^{n+1}}{n+1}$ is equal to $x^n$

My answer is $x^n - x^{n+1} $

Is my algebra at fault? enter image description here

$\endgroup$
  • 2
    $\begingroup$ You're differentiating with respect to $x$. $dn/dx=0$, so your second term is wrong. $\endgroup$ – Chappers Apr 11 '15 at 16:52
  • $\begingroup$ I don't follow what you mean. Can you be a bit more clear? $\endgroup$ – Cetshwayo Apr 11 '15 at 16:53
  • $\begingroup$ n is just a constant. your only variable is x. For example, try differentiating $\frac{1}{2} x^2$ $\endgroup$ – SWilliams Apr 11 '15 at 16:56
  • $\begingroup$ The antiderivative is $\dfrac{x^{n+1}}{n+1}$, not $\dfrac{x^n+1}{n+1}$ $\endgroup$ – egreg Apr 11 '15 at 16:57
9
$\begingroup$

$\dfrac{1}{n+1}$ is a constant, so you can't apply $\dfrac{u}{v}$ rule here like that. Instead, differentiate like this,

$$\frac{\mathrm d}{\mathrm dx}\left(\frac{x^{n+1}}{n+1}\right)=\frac{1}{n+1}\cdot \frac{\mathrm d}{\mathrm dx}\left(x^{n+1}\right)=\frac{n+1}{n+1}x^{n+1-1}=x^n$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Are you using the product rule? The denominator multiplied by the derivative of the numerator? $\endgroup$ – Cetshwayo Apr 11 '15 at 17:01
  • $\begingroup$ @Cetshwayo, Product rule ($uv$ rule) or quotient rule ($\frac{u}{v}$ rule) while differentiating w.r.t. $x$ are applicable when both $u$ and $v$ are functions of $x$ and not some constant. But here, we have a constant. So, we use here the following identity: $$\frac{\mathrm d}{\mathrm dx}(kf(x))=k\cdot \frac{\mathrm d}{\mathrm dx} f(x)~\textrm{where }k\textrm{ is a constant}$$ $\endgroup$ – Prasun Biswas Apr 11 '15 at 17:05
  • 2
    $\begingroup$ Even if you think about product rule or quotient rule, note that $\dfrac{\mathrm d}{\mathrm dx}\left(\dfrac{1}{n+1}\right)=0$, so it doesn't make any difference. $\endgroup$ – Prasun Biswas Apr 11 '15 at 17:09
  • 2
    $\begingroup$ I would like to say that yes you can use the product or quotient rule! There's no restriction saying you can only use it if $u,v$ are non constant. $\endgroup$ – user223391 Apr 11 '15 at 17:24
  • 1
    $\begingroup$ @avid19, ah yes, I noticed that later. Hence, the second comment. :) $\endgroup$ – Prasun Biswas Apr 11 '15 at 17:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.