Let's differentiate: $\frac{\mathrm{d}}{\mathrm{d}x}(x^3-4x)$

ⓘ The derivative of a sum is the sum of the derivatives: $\frac{\mathrm{d}}{\mathrm{d}x}(x^3-4x)$ = $\frac{\mathrm{d}}{\mathrm{d}x}\left(x^3\right)+\frac{\mathrm{d}}{\mathrm{d}x}(-4x)$

$\frac{\mathrm{d}}{\mathrm{d}x}\left(x^3\right)+\frac{\mathrm{d}}{\mathrm{d}x}(-4x)$

ⓘ The derivative of x^n is n*x^(n-1): $\frac{\mathrm{d}}{\mathrm{d}x}\left(x^3\right)$ = $3x^2$■A constant factor stays in front of the derivative, (k*u)' = k*u': $\frac{\mathrm{d}}{\mathrm{d}x}(-4x)$ = $-4$

▶Derivative:   $3x^2-4$