# d/dx[int_1^tanx sqrt(tan^-1t)dt]?

May 3, 2017

By the FTC, Part 1:

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

And by the chain rule:

$\frac{d}{\mathrm{dx}} \left({\int}_{a}^{g \left(x\right)} f \left(t\right) \mathrm{dt}\right) = f \left(g \left(x\right)\right) \frac{\mathrm{dg}}{\mathrm{dx}}$

So:

$\frac{d}{\mathrm{dx}} \left({\int}_{1}^{\tan} x \sqrt{\arctan t} \setminus \mathrm{dt}\right)$

$= \sqrt{\arctan \left(\tan x\right)} \frac{d}{\mathrm{dx}} \left(\tan x\right)$

$= \sqrt{x} {\sec}^{2} x$

May 3, 2017

$\frac{d}{\mathrm{dx}} \setminus {\int}_{1}^{\tan} x \sqrt{{\tan}^{-} 1 t} \setminus \mathrm{dt} = {\sec}^{2} x \sqrt{x}$

#### 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).

$\frac{d}{\mathrm{dx}} \setminus {\int}_{1}^{\tan} x \sqrt{{\tan}^{-} 1 t} \setminus \mathrm{dt}$

(notice the upper and lower bounds 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:

$u = \tan x \implies \frac{\mathrm{du}}{\mathrm{dx}} = {\sec}^{2} x$

The substituting into the integral we get:

$\frac{d}{\mathrm{dx}} \setminus {\int}_{1}^{\tan} x \sqrt{{\tan}^{-} 1 t} \setminus \mathrm{dt} = \frac{d}{\mathrm{dx}} \setminus {\int}_{1}^{u} \sqrt{{\tan}^{-} 1 t} \setminus \mathrm{dt}$
$\text{ } = \frac{\mathrm{du}}{\mathrm{dx}} \cdot \frac{d}{\mathrm{du}} \setminus {\int}_{1}^{u} \sqrt{{\tan}^{-} 1 t} \setminus \mathrm{dt}$
$\text{ } = {\sec}^{2} x \cdot \frac{d}{\mathrm{du}} \setminus {\int}_{1}^{u} \sqrt{{\tan}^{-} 1 t} \setminus \mathrm{dt}$

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

$\frac{d}{\mathrm{du}} \setminus {\int}_{1}^{u} \sqrt{{\tan}^{-} 1 t} \setminus \mathrm{dt} = \sqrt{{\tan}^{-} 1 u}$

And so:

$\frac{d}{\mathrm{dx}} \setminus {\int}_{1}^{\tan} x \sqrt{{\tan}^{-} 1 t} \setminus \mathrm{dt} = {\sec}^{2} x \cdot \sqrt{{\tan}^{-} 1 u}$
$\text{ } = {\sec}^{2} x \cdot \sqrt{{\tan}^{-} 1 \left(\tan x\right)}$
$\text{ } = {\sec}^{2} x \sqrt{x}$