Use the Fundamental Theorem of Calculus to find the derivative of $h(x) = \int_{1}^{e^x} \ln (t) \,dt$ The fundamental theorem of calculus states:

If $f$ is continuous on $[a,b]$, then if $g(x) = \int_{a}^{x}f(t)\,dt,~ \textrm{then}~g'(x) = f(x)$.

In Example 4, the chain rule is used because the upper bound, $x^4$ needed to be differentiated. My question is, why is that and where is that implied given the theorem stated above? 
Similarly, find $h'(x)$ of  $h(x) = \int_{1}^{e^x} \ln(t)\,dt$
 The answer is $h'(x) = xe^x$, which seems to be in line with Example 4. Per previous examples in the book, one might expect the answer to be $h'(x) = \ln(e^x)$ though.

See the proofs of the Leibniz integral rule.
