# How do you use the second fundamental theorem of Calculus to find the derivative of given int cos(t) / t dt  from [3, sqrtx]?

Jul 22, 2017

$\frac{d}{\mathrm{dx}} \setminus {\int}_{3}^{\sqrt{x}} \setminus \cos \frac{t}{t} \setminus \mathrm{dt} = \cos \frac{\sqrt{x}}{2 x}$

#### 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).

We are asked to find:

$\frac{d}{\mathrm{dx}} \setminus = {\int}_{3}^{\sqrt{x}} \setminus \cos \frac{t}{t} \setminus \mathrm{dt}$ ..... [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:

$u = \sqrt{x} \implies \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{2} {x}^{- \frac{1}{2}} = \frac{1}{2 \sqrt{x}}$

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

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

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

$\text{ } = \frac{1}{2 \sqrt{x}} \setminus \frac{d}{\mathrm{du}} \setminus {\int}_{3}^{u} \setminus \cos \frac{t}{t} \setminus \mathrm{dt}$

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

$\frac{d}{\mathrm{dx}} \setminus {\int}_{3}^{\sqrt{x}} \setminus \cos \frac{t}{t} \setminus \mathrm{dt} = \frac{1}{2 \sqrt{x}} \setminus \cos \frac{u}{u}$

And restoring the initial substitution we get:

$\frac{d}{\mathrm{dx}} \setminus {\int}_{3}^{\sqrt{x}} \setminus \cos \frac{t}{t} \setminus \mathrm{dt} = \frac{1}{2 \sqrt{x}} \setminus \cos \frac{\sqrt{x}}{\sqrt{x}}$
$\text{ } = \cos \frac{\sqrt{x}}{2 x}$