True or false: $\frac{\mathrm d}{\mathrm dx}(\int _2^{\:e^x}\:\ln t \,dt)=x-\ln 2$. Support with a proof 
True or false: $\frac{d}{dx}(\int _2^{\:e^x}\:\ln t \,dt)=x-\ln 2$. Support with a proof


I think it is false. I used the Fundamental Theorem of Calculus:
$$\int _2^{\:e^x}\:\ln t\,dt\:=\:\int _0^{e^x}\ln t\:-\:\int _0^2\ln t\,dt$$
The second integral $\frac{d}{dx}\left(\int _0^2\ln t\right)dt=\ln 2$
The first integral $\frac{d}{dx}\left(\int _0^{e^x}\ln t\:\right)=\frac{d\left(e^x\right)}{dx}\frac{d}{d\left(e^x\right)}\int _0^{e^x}\ln t\:=e^xx$ 
Combining the terms $$\frac{d}{dx}\left(\int _2^{\:e^x}\:\ln t\,dt\:\right)=e^xx-\ln 2,$$ 
I saw the solution of the same question on a website, and its answer is different from mine. Is my solution wrong?
 A: Your first step replaces a perfectly normal definite integral with two diverging improper integrals, so you are already on shaky ground.  The logarithm is undefined at zero, so it is not immediate that the integrals on $[0,2]$ and $[0,\mathrm{e}^x]$ are defined.  Far better is to split the interval of integration at a point where the integrand is well-behaved, for instance at $1$.
$$  \int_2^{\mathrm{e}^x} \ln(t) \,\mathrm{d}t =  \int_1^{\mathrm{e}^x} \ln(t) \,\mathrm{d}t - \int_1^{2} \ln(t) \,\mathrm{d}t  $$
This result should make it clear that it doesn't matter whether we split the interval of integration, we are faced with a minor variation of the same integral minus zero.  So don't split. 
Next, the definite integral $\int_1^{2} \ln(t) \,\mathrm{d}t$ is just some number, so $$  \frac{\mathrm{d}}{\mathrm{d}x} \int_1^{2} \ln(t) \,\mathrm{d}t = 0  \text{.}  $$
Finally, let $u(x) = \mathrm{e}^x$ and $f(u) = \int_1^{u} \ln(t) \,\mathrm{d}t$.  Then, \begin{align*}
\frac{\mathrm{d}}{\mathrm{d}x} \int_1^{\mathrm{e}^x} \ln(t) \,\mathrm{d}t
    &= \frac{\mathrm{d}}{\mathrm{d}x} f(u(x))  \\
    &= \frac{\mathrm{d}f}{\mathrm{d}u} \cdot \frac{\mathrm{d}u}{\mathrm{d}x}  \\
    &= \ln(u) \cdot \mathrm{e}^x  \\
    &= x \cdot \mathrm{e}^x  \text{.}  
\end{align*}
A: If you write
$$ F(y) = \int_2^y  \log t dt $$
then as you note ("FTOC") $$ F'(y) = \log y $$
and
$$\begin{align} \frac{d}{dx} F\left(e^x\right) &= F'\left( e^x \right) \frac{d}{dx} e^x \\ &=\log\left( e^x \right) e^x  \\ &= xe^x\end{align}$$
A: $$
\frac{d}{{dx}}\int_2^{e^x } {\log tdt} \mathop  = \limits^{t = e^s } \frac{d}{{dx}}\int_{\log 2}^x {se^s ds}  = xe^x.
$$
