# How do you find F'(x) given F(x)=int 1/t dt from [1,3x]?

Jun 20, 2017

$\frac{d}{\mathrm{dx}} \setminus {\int}_{1}^{3 x} \setminus \frac{1}{t} \setminus \mathrm{dt} = \frac{1}{x}$

#### Explanation:

If asked to find the derivative of an integral then you should not evaluate the integral but instead use the fundamental theorem of Calculus.

The FTOC tells us that:

$\frac{d}{\mathrm{dx}} \setminus {\int}_{a}^{x} \setminus f \left(t\right) \setminus \mathrm{dt} = f \left(x\right)$ for any constant $a$

(ie the derivative of an integral gives us the original function back).

$F ' \left(x\right)$ where $F \left(x\right) = {\int}_{1}^{3 x} \setminus \frac{1}{t} \setminus \mathrm{dt}$

ie, we want:

$F ' \left(x\right) = \frac{d}{\mathrm{dx}} {\int}_{1}^{3 x} \setminus \frac{1}{t} \setminus \mathrm{dt}$ ..... [A}

(notice the upper bounds of the first integral are not in the correct format for the FTOC to be applied, directly). We can manipulate the definite integral using a substitution and the chain rule. Let:

$u = 3 x \implies \frac{\mathrm{du}}{\mathrm{dx}} = 3$

The substituting into the integral [A], and applying the chain rule, we get:

$\frac{d}{\mathrm{dx}} \setminus {\int}_{1}^{3 x} \setminus \frac{1}{t} \setminus \mathrm{dt} = \frac{d}{\mathrm{dx}} \setminus {\int}_{1}^{u} \setminus \frac{1}{t} \setminus \mathrm{dt}$

$\text{ } = \frac{\mathrm{du}}{\mathrm{dx}} \cdot \frac{d}{\mathrm{du}} \setminus {\int}_{1}^{u} \setminus \frac{1}{t} \setminus \mathrm{dt}$

$\text{ } = 3 \setminus \frac{d}{\mathrm{du}} \setminus {\int}_{1}^{u} \setminus \frac{1}{t} \setminus \mathrm{dt}$

And now the derivative of the integral is in the correct form for the FTOC to be applied, giving:

$\frac{d}{\mathrm{dx}} \setminus {\int}_{1}^{3 x} \setminus \frac{1}{t} \setminus \mathrm{dt} = 3 \setminus \frac{1}{u}$

And restoring the initial substitution we get:

$\frac{d}{\mathrm{dx}} \setminus {\int}_{1}^{3 x} \setminus \frac{1}{t} \setminus \mathrm{dt} = 3 \setminus \frac{1}{3 x} = \frac{1}{x}$