OJO: [tex] \frac{d \ ln(u)}{dx} = \frac{u'}{u} [/tex]
=> Si y = ln(x^2 + 5) , tendriamos que:
[tex] \frac{d \ y}{dx} = y' = \frac{(x^2+5)'}{x^2+5}
\ \
[/tex]
[tex]
\frac{d \ y}{dx} = y' = \frac{2x^{2-1}}{x^2+5} [/tex]
[tex]
\frac{d \ y}{dx} = y' = \frac{2x}{x^2+5} [/tex] ← Respuesta
Eso es todo!