To solve this problem, we need to apply the Leibniz rule for differentiating under the integral sign. The Leibniz rule states that if you have an integral of the form \(\int_{a(x)}^{b(x)} f(t) \, dt\), the derivative with respect to \(x\) is given by:
\[
\frac{d}{dx} \left[ \int_{a(x)}^{b(x)} f(t) \, dt \right] = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x)
\]
In this case, \(a(x) = 2\) and \(b(x) = \tan x\), and the function \(f(t) = \frac{1}{1+t^2}\). Since \(a(x)\) is a constant, its derivative is zero. We need to evaluate \(f(b(x))\) and multiply it by the derivative of \(b(x)\).
Para resolver \(\frac{d}{dx}\left[\int_{2}^{\tan x} \frac{1}{1+t^{2}} \, dt\right]\), utilizamos la regla de Leibniz para la diferenciación bajo el signo integral. La regla establece que:
\[
\frac{d}{dx} \left[ \int_{a(x)}^{b(x)} f(t) \, dt \right] = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x)
\]
En este caso, \(a(x) = 2\) y \(b(x) = \tan x\), y la función \(f(t) = \frac{1}{1+t^2}\).
Sustituimos \(t = \tan x\) en \(f(t)\):
\[
f(\tan x) = \frac{1}{1+(\tan x)^2}
\]
La derivada de \(b(x) = \tan x\) es:
\[
b'(x) = \sec^2 x
\]
Dado que \(a(x) = 2\) es constante, su derivada es cero. Por lo tanto, la derivada es:
\[
f(\tan x) \cdot b'(x) = \frac{1}{1+(\tan x)^2} \cdot \sec^2 x
\]
Utilizando la identidad trigonométrica \(\sec^2 x = 1 + \tan^2 x\), simplificamos:
\[
\frac{1}{1+(\tan x)^2} \cdot (1 + \tan^2 x) = 1
\]
La derivada de la integral dada es:
\[
\boxed{1}
\]
Answered
