What is the derivative of $\frac{x^{n+1}}{n+1}$?

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?

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

$\dfrac{1}{n+1}$ is a constant, so you can't apply $\dfrac{u}{v}$ rule here like that. Instead, differentiate like this,
$$\frac{\mathrm d}{\mathrm dx}\left(\frac{x^{n+1}}{n+1}\right)=\frac{1}{n+1}\cdot \frac{\mathrm d}{\mathrm dx}\left(x^{n+1}\right)=\frac{n+1}{n+1}x^{n+1-1}=x^n$$
• @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: $$\frac{\mathrm d}{\mathrm dx}(kf(x))=k\cdot \frac{\mathrm d}{\mathrm dx} f(x)~\textrm{where }k\textrm{ is a constant}$$ – Prasun Biswas Apr 11 '15 at 17:05
• 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. – Prasun Biswas Apr 11 '15 at 17:09
• 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. – user223391 Apr 11 '15 at 17:24