Quick integral question Sorry about the formatting, but how would I go about this question:
$$\frac{d}{dx} \int_{\cos x}^1 \sqrt{(1 + e)^t} dt$$
What I've learned in class is that the derivative of an integral is just the function itself ($\sqrt{(1 + e)^t}$)
However, it doesn't seem to be the case here.
Thanks for your time
 A: Using the Newton-Leibnitz rule,
$$\frac{d}{dx} \int_{f(x)}^{g(x)} h(t) dt=h(g(x))g'(x)-h(f(x))f'(x)$$
This should solve the problem.
A: Use the Fundamental Theorem of Calculus:
$$\frac{d}{dx} \int_{g(x)}^a f(t) dt = \frac{d}{dx}\big( F(a) - F(g(x)) \big)= -g'(x) f(g(x))$$
So in your case:
$$\frac{d}{dx} \int_{\cos x}^1 \sqrt{(1 + e)^t} dt = \sin(x)\sqrt{(1 + e)^{\cos x}}$$
A: Here you have Integration Limits that depend on $x$. In this case, the Differentiation of the integral is slightly modified. Use the Leibnitz rule for Integration. 
A: Notice, $$\frac {d}{dx}\int_{\cos x}^1\sqrt{(1+e)^{t}}\ dt$$
$$=\frac {d}{dx}\int_{\cos x}^1(1+e)^{t/2}\ dt$$
using Fundamental Theorem of Calculus (F.T.C.), 
$$=(1+e)^{\frac{1}{2}}\frac {d}{dx}(1)-(1+e)^{\Large \frac{\cos x}{2}}\frac {d}{dx}(\cos x)$$
$$=0-(1+e)^{\Large \frac{\cos x}{2}}(-\sin x)$$
$$=\color{blue}{\sin x(1+e)^{\Large \frac{\cos x}{2}}}$$
A: Let $h(t) = \sqrt{(1+e)^t}$, and let
$$
y = \int_{\cos{x}}^{1} h(t) dt =  \int_{1}^{\cos x} -h(t) dt.
$$
You want to find $\frac{dy}{dx}$.
The problem is the top endpoint, which is $\cos x$.  If it were just an $x$, then you'd know exactly how to do it--just apply the Fundamental Theorem of Calculus and you'd be done.
So let's make a quick substitution: write $u = \cos x$.  Now
$$
y = \int_{1}^{u} -h(t) dt.
$$
So by the FTC, we now know that $\frac{dy}{du} = -h(u)$.  But we didn't want $\frac{dy}{du}$, we wanted $\frac{dy}{dx}$.
Well, remember the Chain Rule:
$$
\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}
$$
So you just need to find $\frac{du}{dx}$ and you'll be done.  (Remember to write your answer in terms of $x$, the original variable.)
