Applicare la formula: $\frac{d}{dx}\left(a^b\right)$$=y=a^b$, dove $d/dx=\frac{d}{dx}$, $a=x^2+3$, $b=5x-1$, $a^b=\left(x^2+3\right)^{\left(5x-1\right)}$ e $d/dx?a^b=\frac{d}{dx}\left(\left(x^2+3\right)^{\left(5x-1\right)}\right)$

Applicare la formula: $y=a^b$$\to \ln\left(y\right)=\ln\left(a^b\right)$, dove $a=x^2+3$ e $b=5x-1$

Applicare la formula: $\ln\left(x^a\right)$$=a\ln\left(x\right)$, dove $a=5x-1$ e $x=x^2+3$

Applicare la formula: $\ln\left(y\right)=x$$\to \frac{d}{dx}\left(\ln\left(y\right)\right)=\frac{d}{dx}\left(x\right)$, dove $x=\left(5x-1\right)\ln\left(x^2+3\right)$

Applicare la formula: $\frac{d}{dx}\left(ab\right)$$=\frac{d}{dx}\left(a\right)b+a\frac{d}{dx}\left(b\right)$, dove $d/dx=\frac{d}{dx}$, $ab=\left(5x-1\right)\ln\left(x^2+3\right)$, $a=5x-1$, $b=\ln\left(x^2+3\right)$ e $d/dx?ab=\frac{d}{dx}\left(\left(5x-1\right)\ln\left(x^2+3\right)\right)$

Applicare la formula: $\frac{d}{dx}\left(x\right)$$=1$

Applicare la formula: $\frac{d}{dx}\left(x\right)$$=1$

Applicare la formula: $\frac{a}{b}=c$$\to a=cb$, dove $a=y^{\prime}$, $b=y$ e $c=5\ln\left(x^2+3\right)+\frac{2\left(5x-1\right)x}{x^2+3}$

Sostituire $y$ con la funzione originale: $\left(x^2+3\right)^{\left(5x-1\right)}$

La derivata della funzione risulta