If $ \displaystyle F(x) = \int_{1}^{3 x} \frac{1}{t} ~ \mathrm{d}{t} $, find $ F'(x) $. Did I do the following correctly?
\begin{align*}
    F'(x)
& = \frac{\mathrm{d}}{\mathrm{d}{x}}
    \left[ \int_{1}^{3 x} \frac{1}{t} ~ \mathrm{d}{t} \right] \\
& = \frac{\mathrm{d}}{\mathrm{d}{x}} [\ln(3 x) - \ln(1)] \\
& = \frac{\mathrm{d}}{\mathrm{d}{x}} [\ln(3 x)] \\
& = \frac{1}{x}.
\end{align*}
 A: Yes, it is definitely correct. You can also do it the following way, using the Fundamental Theorem of Calculus:
\begin{align}
F'(x) &= \frac{d}{dx} \left[ \int_{1}^{3x} \frac{1}{t} \,d{t} \right] \\
      &= \frac{d}{du} \left[ \int_{1}^{u} \frac{1}{t} \,d{t} \right] \cdot \frac{du}{dx} \quad (\text{Using the substitution $ u = 3x $ and the Chain Rule.}) \\
      &= \frac{1}{u} \cdot 3 \quad (\text{By the Fundamental Theorem of Calculus.}) \\
      &= \frac{1}{3x} \cdot 3 \\
      &= \frac{1}{x}.
\end{align}
This method works even if you have a more complicated integrand than $ \dfrac{1}{t} $. It is what I usually show my students in my freshman calculus class after I have taught them the Fundamental Theorem of Calculus.
A: Yes. Your answer is indeed correct. However, you need not have performed the integral. You could resort to Leibniz integral rule (Fundamental theorem of Calculus in disguise) for this. The rule goes as follows.
If $$F(x) =  \int_{a(x)}^{b(x)} f(x,t) dt,$$ then $$F'(x) = \int_{a(x)}^{b(x)} \dfrac{\partial f(x,t)}{\partial x} dt + f(x,b(x)) \dfrac{db(x)}{dx} - f(x,a(x)) \dfrac{da(x)}{dx}$$
In your case, $f(x,t) = \dfrac1t$, $a(x)=1$ and $b(x) = 3x$. Hence, we find that the first and last terms in the above equality are zero. The only non-zero term is the second term i.e.
$$F'(x) = f(x,b(x)) \dfrac{db(x)}{dx} = \dfrac1{3x} \times 3 = \dfrac1x$$
Leibniz integral rule is especially handy when integrating $f(x,t)$ i.e. computing $\displaystyle \int_{a(x)}^{b(x)} f(x,t) dt$ explicitly is difficult.
A: You should google "Differentiation under the Integral sign".
Define a function:
$$G(x)=\int_1^x \frac {1}{t} \mathrm{d}t$$
So,
$$F(x)=G(3x) \to F'(x)=[G(3x)]'=3G'(3x)=3\cdot \frac {1}{3x}=\frac{1}{x} $$
Where I used the chain rule for differentiation, and the fundamental theorem of calculus.
