Find the derivative of $f(x)= \int_{\sin x}^{\tan x} \sqrt{t^{2}+t+1}\, \mathrm d t$ 
Find the derivative of $$f(x)=\int_{\sin x}^{\tan x} \sqrt{t^{2}+t+1}\, 
 \mathrm d t$$ with respect to $x$

So from may understanding, I need to apply the fundamental theorem of calculus and then differentiate. I think the upper and lower limits are throwing me off.

Let $g(t)$ denote the integrand $\sqrt{t^2+t+1}$. On the one hand, the FTC guarantees
$$
\frac{d}{dx}\int_{\sin x}^{\tan x} g(t)\, dt
$$
$$
=g(\tan(x))\cdot (\tan(x))' - g(\sin(x))\cdot (\sin(x))'
$$
$$
=g(\tan(x))\cdot \sec^2(x) - g(\sin(x))\cdot \cos(x)
$$
$$
=\sqrt{\tan^{2} (x)+\tan(x)+1}\cdot \sec^2(x) - \sqrt{\sin^{2} (x)+\sin(x)+1}\cdot \cos(x)
$$Were we masochisitic, we could compute the antiderivative using the substitution $(t+1/2)^2= (3/4)\tan^2(\theta)$ (note this cannot always be done, which is part of the power of the FTC), back-substitute, and then differentiate to verify we get the same result.
$$
\int \sqrt{t^2+t+1}\,dt = \int \sqrt{(t+1/2)^2+3/4}\,dt
$$
$$
=\frac{1}{2} t\sqrt{t^2+t+1} +\frac{1}{4} \sqrt{t^2+t+1}+\frac{3}{8} \log \left(\frac{2
   t+1}{\sqrt{3}}+\sqrt{\frac{1}{3} (2 t+1)^2+1}\right)
$$For instance, replacing $t$ with $\tan(x)$ at the upper limit and differentiating gives:
$$
\frac{d}{dx}\left(\frac{1}{2} \tan (x) \sqrt{\tan ^2(x)+\tan (x)+1}+\frac{1}{4} \sqrt{\tan ^2(x)+\tan
   (x)+1}+\frac{3}{8} \log \left(\frac{2 \tan (x)+1}{\sqrt{3}}+\sqrt{\frac{1}{3} (2 \tan
   (x)+1)^2+1}\right)\right)
$$
$$
=\frac{1}{2} \sqrt{\tan ^2(x)+\tan (x)+1} \sec ^2(x)+\frac{\tan (x) \left(\sec ^2(x)+2 \tan (x)
   \sec ^2(x)\right)}{4 \sqrt{\tan ^2(x)+\tan (x)+1}}+\frac{\sec ^2(x)+2 \tan (x) \sec ^2(x)}{8
   \sqrt{\tan ^2(x)+\tan (x)+1}}+\frac{3 \left(\frac{2 \sec ^2(x)}{\sqrt{3}}+\frac{2 (2 \tan
   (x)+1) \sec ^2(x)}{3 \sqrt{\frac{1}{3} (2 \tan (x)+1)^2+1}}\right)}{8 \left(\frac{2 \tan
   (x)+1}{\sqrt{3}}+\sqrt{\frac{1}{3} (2 \tan (x)+1)^2+1}\right)}$$
$$
=\sqrt{\tan^2(x)+\tan(x)+1}\cdot \sec^2(x),
$$as promised. If you want, you can try the lower limit.
