Derivative of $\int x^5 (x^6 - 6)^4 \rm dx$ Find the derivative of $\int x^5 (x^6 - 6)^4\, \rm dx$
I'm not sure how to do this problem with the integral. If you could provide a thorough explanation, that would be great. 
 A: The derivate is simply the function under the integral (fundamental theorem of calculus) 
A: If you are to find the derivative of $$\int_0^t x^5(x^6 - 6)^4 \,dx$$ then use the Fundamental Theorem of Calculus.
If you are to evaluate the given integral, then use $u$-substitution by letting $$\color{blue}{\bf u = x^6 - 6}\implies du = 6x^5\,dx \iff \,\color{red}{\bf x^5dx = \dfrac {du}6}$$
$$\int \color{red}{\bf x^5}(\color{blue}{\bf x^6 - 6})^4 \color{red}{\bf \,dx} = \int \color{blue}{\bf u}^4\,\color{red}{\bf \dfrac{du}6} = \dfrac 16\int u^4 \,du$$
Now use the power rule to integrate: $$\int u^n\,du = \dfrac{u^{n+1}}{n+1} + C,\;\;n\neq -1$$
A: Fundamental Theorem of Calculus, Part I:
Let f be a continuous real-valued function defined on a closed interval $[a,b]$. Let $F$ eb the function defined, for all $x \in [a,b]$, by
$$F(x) = \int^x_a f(t) dt $$.
Then F is continuous on $[a,b]$, differentiable on $(a,b)$, and 
$$F'(x)=f(x)$$ for all $x \in (a,b)$.
So in your case $F(x):=\displaystyle \int^x_a  t^5 (t^6 - 6)^4 dt = \int^x_a f(t) dt$, and so as $F'(x)=f(x)$, we get that $F'(x)=x^5(x^6-6)^4$.
$a$ can be anything as it will only cause constant terms which will equal zero after the derivative is taken.
