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Derivado de xxx^{x}

La calculadora hallará la derivada de xxx^{x}, con los pasos mostrados.

Calculadora relacionada: Calculadora de derivadas

Solución

Utilice la fórmula fg(x)(x)=eg(x)ln⁡(f(x))f^{g{\left(x \right)}}{\left(x \right)} = e^{g{\left(x \right)} \ln\left(f{\left(x \right)}\right)} con f(x)=xf{\left(x \right)} = x y g(x)=xg{\left(x \right)} = x para reescribir la expresión compleja:

(ddx(xx))=(ddx(exln⁡(x))){\color{red}\left(\frac{d}{dx} \left(x^{x}\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(e^{x \ln\left(x\right)}\right)\right)}

La función exln⁡(x)e^{x \ln\left(x\right)} es la composición f(g(x))f{\left(g{\left(x \right)} \right)} de dos funciones f(u)=euf{\left(u \right)} = e^{u} y g(x)=xln⁡(x)g{\left(x \right)} = x \ln\left(x\right).

Aplique la regla de la cadena ddx(f(g(x)))=ddu(f(u))ddx(g(x))\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right):

(ddx(exln⁡(x)))=(ddu(eu)ddx(xln⁡(x))){\color{red}\left(\frac{d}{dx} \left(e^{x \ln\left(x\right)}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(e^{u}\right) \frac{d}{dx} \left(x \ln\left(x\right)\right)\right)}

La derivada de la exponencial es ddu(eu)=eu\frac{d}{du} \left(e^{u}\right) = e^{u}:

(ddu(eu))ddx(xln⁡(x))=(eu)ddx(xln⁡(x)){\color{red}\left(\frac{d}{du} \left(e^{u}\right)\right)} \frac{d}{dx} \left(x \ln\left(x\right)\right) = {\color{red}\left(e^{u}\right)} \frac{d}{dx} \left(x \ln\left(x\right)\right)

Volver a la antigua variable:

e(u)ddx(xln⁡(x))=e(xln⁡(x))ddx(xln⁡(x))=xxddx(xln⁡(x))e^{{\color{red}\left(u\right)}} \frac{d}{dx} \left(x \ln\left(x\right)\right) = e^{{\color{red}\left(x \ln\left(x\right)\right)}} \frac{d}{dx} \left(x \ln\left(x\right)\right) = x^{x} \frac{d}{dx} \left(x \ln\left(x\right)\right)

Aplica la regla del producto ddx(f(x)g(x))=ddx(f(x))g(x)+f(x)ddx(g(x))\frac{d}{dx} \left(f{\left(x \right)} g{\left(x \right)}\right) = \frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} + f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right) con f(x)=xf{\left(x \right)} = x y g(x)=ln⁡(x)g{\left(x \right)} = \ln\left(x\right):

xx(ddx(xln⁡(x)))=xx(ddx(x)ln⁡(x)+xddx(ln⁡(x)))x^{x} {\color{red}\left(\frac{d}{dx} \left(x \ln\left(x\right)\right)\right)} = x^{x} {\color{red}\left(\frac{d}{dx} \left(x\right) \ln\left(x\right) + x \frac{d}{dx} \left(\ln\left(x\right)\right)\right)}

La derivada del logaritmo natural es ddx(ln⁡(x))=1x\frac{d}{dx} \left(\ln\left(x\right)\right) = \frac{1}{x}:

xx(x(ddx(ln⁡(x)))+ln⁡(x)ddx(x))=xx(x(1x)+ln⁡(x)ddx(x))x^{x} \left(x {\color{red}\left(\frac{d}{dx} \left(\ln\left(x\right)\right)\right)} + \ln\left(x\right) \frac{d}{dx} \left(x\right)\right) = x^{x} \left(x {\color{red}\left(\frac{1}{x}\right)} + \ln\left(x\right) \frac{d}{dx} \left(x\right)\right)

Aplique la regla de potencia ddx(xn)=nxn−1\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1} con n=1n = 1, es decir, ddx(x)=1\frac{d}{dx} \left(x\right) = 1:

xx(ln⁡(x)(ddx(x))+1)=xx(ln⁡(x)(1)+1)x^{x} \left(\ln\left(x\right) {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} + 1\right) = x^{x} \left(\ln\left(x\right) {\color{red}\left(1\right)} + 1\right)

Así, ddx(xx)=xx(ln⁡(x)+1)\frac{d}{dx} \left(x^{x}\right) = x^{x} \left(\ln\left(x\right) + 1\right).

Respuesta

ddx(xx)=xx(ln⁡(x)+1)\frac{d}{dx} \left(x^{x}\right) = x^{x} \left(\ln\left(x\right) + 1\right)A