Question

Să se calculeze derivata: f(x)=x^(x+1)

Answer 1

f(x)=xˣ⁺¹
d/dx(f(x))=d/dx(x¹⁺ˣ)
f°(x)=d/dx(x¹⁺ˣ)=d/dx[e^((1+x)ln(x))]=d/dx((1+x)ln(x))*e^[(1+x)ln(x)]=
=xˣ⁺¹d/dx((1+x)ln(x))=ln(x)d/dx(1+x)+(x+1)d/dx(ln(x))x¹⁺ˣ=
=x¹⁺ˣ[ln(x)(d/dx(x))+(1+x)(d/dx(ln(x)))]=
=x¹⁺ˣ(ln(x)+(1+x)1/x)=
=x¹⁺ˣ[(1+x)/x+ln(x)]

f°(x)=xˣ(x+xln(x)+1)

Answer 2

[tex]f'(x)=(x^{x+1})'=(e^{ln(x^{x+1})})'=(e^{(x+1)lnx})'=e^{(x+1)lnx}\cdot((x+1)lnx)'\\ =x^{x+1}\cdot (lnx+ \frac{x+1}{x} )=\\ =x^{x+1}\cdot lnx+ x^{x}\cdot (x+1)=\\ =x^x(xlnx+x+1)[/tex]