$F(x) =\int \limits_{\tan x}^{\cot x}\sqrt{1+t^2}\,\mathrm{d}t\,\,$ then $F'( \pi/4) =?$ 
$F(x) =\int \limits_{\tan x}^{\cot x}\sqrt{1+t^2}\,\mathrm{d}t\,\,$
  then $F'(\pi/4) =?$

$$F(x) =\int \limits_{\tan x}^{\cot x}\sqrt{1+t^2}\,\mathrm{d}t$$
$$F(x) =-\int \limits_{0}^{\cot x}\sqrt{1+t^2}\,\mathrm{d}t + \int \limits_{0}^{\tan x}\sqrt{1+t^2}\,\mathrm{d}t$$
According to Fundamental Theorem Part 1:
\begin{align}Old\\
F'(x) & = f(b(x))\cdot b'(x)-f(a(x))\cdot a'(x) \\
F'(x) & = \sqrt{1+x^2}(-\csc^2x) - \sqrt{1+x^2}(\sec^2x) \\
F'(x) & = -\sqrt{1+x^2}(\csc^2x + \sec^2x) \\
F'(x) & = -\sqrt{1+x^2}(1) \\
F'(\pi/4) & = -\sqrt{1+(\pi/4)^2} \\
\end{align}
New 
$F'(x)  = \sqrt{1+x^2}(d/dx(\sqrt{1+\cot^2x}) - \sqrt{1+x^2}(d/dx(\sqrt{1+\tan^2x})$
$F'(x)  = \sqrt{1+x^2}(d/dx(\sqrt{\csc^2x}) - \sqrt{1+x^2}(d/dx(\sqrt{\sec^2x})$
$F'(x)  = \sqrt{1+x^2}[(-\csc{x}\cot{x}) - (-\sec{x}\tan{x}))$
$F'(\pi/4)  = \sqrt{1+(\pi/4)^2}[(-\csc{(\pi/4)}\cot{(\pi/4)}) - (\sec{(\pi/4)}\tan{(\pi/4)}))$
?
 A: Hint: If $F(x) = \int_{a(x)}^{b(x)}f(t)dt$, then by the fundamental theorem of calculus and chain rule:

$$
F'(x)=f(b(x))b'(x)-f(a(x))a'(x)
$$


To see why this works, observe that:
$$ \begin{align*}
\int_{a(x)}^{b(x)}f(t)dt &= \int_{a(x)}^{0}f(t)dt + \int_{0}^{b(x)}f(t)dt = \int_{0}^{b(x)}f(t)dt - \int_{0}^{a(x)}f(t)dt \\
\end{align*} $$
Now let $u=a(x)$ and $v=b(x)$. Then we have:
$$ \begin{align*}
F'(x) &= \dfrac{d}{dx} \left[ \int_{0}^{b(x)}f(t)dt \right] - \dfrac{d}{dx} \left[ \int_{0}^{a(x)}f(t)dt \right] \\
&= \dfrac{d}{dx} \left[ \int_{0}^{v}f(t)dt \right] - \dfrac{d}{dx} \left[ \int_{0}^{u}f(t)dt \right] \\
&= f(v) \dfrac{dv}{dx} - f(u) \dfrac{du}{dx} \\
&= f(b(x)) b'(x) - f(a(x)) a'(x) \\
\end{align*} $$

Edit: Thanks for editing in the work that you have so far. You're almost there; note that $f(b(x)) = \sqrt{1+\cot^2x} = \sqrt{\csc^2x}$ and $f(a(x))=\sqrt{1+\tan^2x} = \sqrt{\sec^2x}$.

Edit 2: Since $a(x)=\tan x$ and $b(x)=\cot x$, we have $a'(x)=\sec^2x$ and $b'(x)=-\csc^2x$. Putting everything together, our first step should be:
$$
F'(x)= \left(\sqrt{1+[b(x)]^2} \right)(-\csc^2x) - \left(\sqrt{1+[a(x)]^2}\right)(\sec^2x)
$$
