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Did I do this correctly?

$$F^{\prime}(x) = \frac{\mathrm{d}}{\mathrm{d}x} \int_1^{3x} \frac{1}{t}\, \mathrm{d}t $$

$$ \Rightarrow \frac{\mathrm{d}}{\mathrm{d}x} \left[ \ln (3x) - \ln (1) \right] $$ $$ \Rightarrow \frac{\mathrm{d}}{\mathrm{d}x} \left[ \ln (3x) \right]$$ $$ \Rightarrow \frac{1}{x} $$

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Yes, although there is no need to know the anti-derivative explicitly. Use the chain rule: $$ \frac{\mathrm{d}}{\mathrm{d}x} \int_1^{3x} \frac{1}{t} \mathrm{d} t = \frac{1}{3 x} \frac{\mathrm{d} (3x)}{\mathrm{d}x} = \frac{1}{x} $$ – Sasha Nov 21 '12 at 5:04
up vote 4 down vote accepted

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.

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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.

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Based on the level of the OP's question, I doubt that he/she is familiar with partial derivatives, or would know what the notation $\partial f(x,t)/\partial x$ means. – Jesse Madnick Nov 21 '12 at 5:26
@JesseMadnick But I learn more from Marvis's answer, Marvis's answer gave me something more to read and learn about, other than the simple terminal answers mostly given here. – yiyi Nov 22 '12 at 0:39
I stand happily corrected :-) – Jesse Madnick Nov 22 '12 at 1:09

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.

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+1. You should change $G(3y)$ to $G(3x)$, however. – JavaMan Nov 21 '12 at 5:27
edited it - thx. – Amihai Zivan Nov 21 '12 at 5:43

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