Let's differentiate: $\frac{\mathrm{d}}{\mathrm{d}x}(x^5+3x)$

ⓘ The derivative of a sum is the sum of the derivatives: $\frac{\mathrm{d}}{\mathrm{d}x}(x^5+3x)$ = $\frac{\mathrm{d}}{\mathrm{d}x}\left(x^5\right)+\frac{\mathrm{d}}{\mathrm{d}x}\left(3x\right)$

$\frac{\mathrm{d}}{\mathrm{d}x}\left(x^5\right)+\frac{\mathrm{d}}{\mathrm{d}x}\left(3x\right)$

ⓘ The derivative of x^n is n*x^(n-1): $\frac{\mathrm{d}}{\mathrm{d}x}\left(x^5\right)$ = $5x^4$■A constant factor stays in front of the derivative, (k*u)' = k*u': $\frac{\mathrm{d}}{\mathrm{d}x}\left(3x\right)$ = $3$

▶Derivative:   $5x^4+3$