Let's differentiate: $\frac{\mathrm{d}}{\mathrm{d}x}(x^2+2x+1)$

ⓘ The derivative of a sum is the sum of the derivatives: $\frac{\mathrm{d}}{\mathrm{d}x}(x^2+2x+1)$ = $\frac{\mathrm{d}}{\mathrm{d}x}\left(x^2\right)+\frac{\mathrm{d}}{\mathrm{d}x}\left(2x\right)+0$

$\frac{\mathrm{d}}{\mathrm{d}x}\left(x^2\right)+\frac{\mathrm{d}}{\mathrm{d}x}\left(2x\right)+0$

ⓘ The derivative of x^n is n*x^(n-1): $\frac{\mathrm{d}}{\mathrm{d}x}\left(x^2\right)$ = $2x$■Adding 0 changes nothing: $\frac{\mathrm{d}}{\mathrm{d}x}\left(2x\right)+0$ = $\frac{\mathrm{d}}{\mathrm{d}x}\left(2x\right)$■A constant factor stays in front of the derivative, (k*u)' = k*u': $\frac{\mathrm{d}}{\mathrm{d}x}\left(2x\right)$ = $2$

▶Derivative:   $2x+2$