# 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!

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Unfortunately, it is the second answer which is right (the one with $-6x^5$). –  André Nicolas Dec 6 '11 at 20:14
Indeed, by the FTC and the chain rule, it's the latter. –  The Chaz 2.0 Dec 6 '11 at 20:14
Yeah I had the feeling I did the chain rule wrong thanks though sigh –  joe Dec 6 '11 at 20:15
You should post your answer. –  David Mitra Dec 6 '11 at 20:33

## 3 Answers

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.

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"This answer is useful" - especially for those who aren't allowed to access W|A during tests. –  The Chaz 2.0 Dec 7 '11 at 3:13

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.

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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:

Note that the latter can already be achieved by just pressing the result in the first link.

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