To find the derivative of the given function, we need to apply the chain rule, product rule, and implicit differentiation. The function is a sum of two terms: \(\sin(x)^{\cos(x)}\) and \((\ln(x))^{\int_{1}^{x^{2}} \frac{1}{\tan(t)+1} dt}\). For the first term, use logarithmic differentiation to handle the power of a function. For the second term, differentiate the integral with respect to \(x\) using the Fundamental Theorem of Calculus and then apply the chain rule.
Para el término \(\sin(x)^{\cos(x)}\), utilizamos la diferenciación logarítmica. Sea \(y = \sin(x)^{\cos(x)}\), entonces \(\ln(y) = \cos(x) \ln(\sin(x))\). Derivando ambos lados con respecto a \(x\), obtenemos:
\[
\frac{1}{y} \frac{dy}{dx} = -\sin(x) \ln(\sin(x)) + \frac{\cos(x)^2}{\sin(x)}
\]
Por lo tanto, la derivada del primer término es:
\[
\frac{dy}{dx} = \left(-\ln(\sin(x)) \sin(x) + \frac{\cos(x)^2}{\sin(x)}\right) \sin(x)^{\cos(x)}
\]
Para el término \((\ln(x))^{\int_{1}^{x^{2}} \frac{1}{\tan(t)+1} dt}\), primero calculamos la derivada del exponente usando el teorema fundamental del cálculo:
\[
\frac{d}{dx} \left(\int_{1}^{x^2} \frac{1}{\tan(t) + 1} dt\right) = \frac{d}{dx} \left(\frac{x^2}{2} + \frac{\ln(\tan(x^2) + 1)}{2} - \frac{\ln(\tan(x^2)^2 + 1)}{4} - \frac{1}{2} - \frac{\ln(1 + \tan(1))}{2} + \frac{\ln(1 + \tan(1)^2)}{4}\right)
\]
La derivada es:
\[
x - x \tan(x^2) + \frac{x (\tan(x^2)^2 + 1)}{\tan(x^2) + 1}
\]
Ahora, aplicamos la regla de la cadena para el término completo:
\[
\frac{d}{dx} \left((\ln(x))^{\text{exponente}}\right) = \left(\text{exponente} \cdot \frac{1}{x \ln(x)} + \ln(\ln(x)) \cdot \text{derivada del exponente}\right) \cdot (\ln(x))^{\text{exponente}}
\]
La derivada completa de la función dada es:
\[
\boxed{\left((-x \tan(x^2) + x + \frac{x (\tan(x^2)^2 + 1)}{\tan(x^2) + 1}) \ln(\ln(x)) + \frac{\text{exponente}}{x \ln(x)}\right) (\ln(x))^{\text{exponente}} + \left(-\ln(\sin(x)) \sin(x) + \frac{\cos(x)^2}{\sin(x)}\right) \sin(x)^{\cos(x)}}
\]
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