Need help with derivative of an integral I was wondering what the answer to this question is. I just had a test and I want to make sure I was correct. So I need to find the derivative of 
$$
\int_{x^6}^{0} \cos(\sqrt{t}) ~ dt
$$
(I believe it is $dt$ however I may be wrong and it could have been $dx$.) 
The answer I thought was $-3x^2 \cos(x^3)$ but now I realize the answer might be 
$-6x^5 \cos(x^3)$. Thank you in advance!
 A: Here's one way to do this kind of problem (without Wolfram). 
Let $F(t)=\int\cos(\sqrt t)\,dt$, so $$F'(t)=\cos(\sqrt t)$$ Then $$\int_{x^6}^0\cos(\sqrt t)\,dt=F(0)-F(x^6)$$ So the derivative is $(F(0))'-(F(x^6))'$. Well, $F(0)$ is a constant, so its derivative is zero, so we just have to figure out $-(F(x^6))'$. By the chain rule, this is  $$-F'(x^6)(x^6)'=-\cos(\sqrt{x^6})(6x^5)=-6x^5\cos|x^3|=-6x^5\cos(x^3)$$ since cosine is an even function. 
A: By the Fundamental Theorem of Calculus and the chain rule, 
If $F(x) = \int_{a}^{g(x)}f(t)dt$, then $F\ '(x) = f(g(x)) \cdot g'(x)$.
In particular, for $f(t) = \cos(\sqrt{t}); g(x) = x^6$, we have $F \ '(x) = 6x^5 \cdot \cos(x^3)$ 
(with simplifications as in Gerry Myerson's answer). 
But then we have that $\int_a^b \text{(stuff)} = - \int_b^a \text{(stuff)}$, which gives us the "$-$" in front as desired. 
A: As you correctly realized the right answer is $-6x^5\cos(x^3)$.
However, just for confirming such results you don't need to ask us, there are efficient means to help yourself:


*

*Use Wolframalpha to integrate your expression

*Use it to derive your expression
Note that the latter can already be achieved by just pressing the result in the first link.
