To prove $\int x^p \, dx = \frac{x^{p+1}}{p+1} + C$, my calculus textbook writes:
$$F '(x) = \frac{d}{dx} \left(\frac{x^{p+1}}{p+1} +C\right) = \frac{d}{dx} \left(\frac{x^{p+1}}{p+1}\right)+\frac{d}{dx}(C)=\frac{(p+1)x^p}{p+1}+0=x^p.$$
I am confused on how they take the derivative of $\dfrac{x^p+1}{p+1}$ without using the quotient rule. Can someone please explain to me why it is that they apply the power rule to the numerator but seem to completely ignore the denominator?
On
By definition, $$ {d \over dx}x^n = \lim_{h \rightarrow 0} {(x+h)^n - x^n \over h}. $$ Now, by Newton's binomial theorem, $$ (x + h)^n = x^n + n h x^{n-1} + (\mbox{2nd and higher powers of $h$}). $$ Subtracting $x^n$ and dividing by $h$, we get $$ {(x+h)^n - x^n \over h} = n x^{n-1} + (\mbox{1st and higher powers of $h$}) $$ Now, letting $h \rightarrow 0$ on the right-hand side leaves just $nx^{n-1}$.
$p$ is a constant, so $\frac{1}{p+1}$ is a constant, so $\frac{x^{p+1}}{p+1}=\frac{1}{p+1}x^{p+1}$. Just because there's a quotient doesn't mean we need to think of it as a quotient.