Let's perform the differentiation: $\frac{\mathrm{d}}{\mathrm{d}x}(5x^2-1)$

ⓘ The derivative of a sum is the sum of the derivatives: $\frac{\mathrm{d}}{\mathrm{d}x}(5x^2-1)$ = $\frac{\mathrm{d}}{\mathrm{d}x}\left(5x^2\right)+0$

$\frac{\mathrm{d}}{\mathrm{d}x}\left(5x^2\right)+0$

ⓘ Adding 0 changes nothing: $\frac{\mathrm{d}}{\mathrm{d}x}\left(5x^2\right)+0$ = $\frac{\mathrm{d}}{\mathrm{d}x}\left(5x^2\right)$■A constant factor stays in front of the derivative, (k*u)' = k*u': $\frac{\mathrm{d}}{\mathrm{d}x}\left(5x^2\right)$ = $5·\frac{\mathrm{d}}{\mathrm{d}x}\left(x^2\right)$■The derivative of x^n is n*x^(n-1): $\frac{\mathrm{d}}{\mathrm{d}x}\left(x^2\right)$ = $2x$

$5·2x$

ⅰ $5·2$ = $10$

▶Derivative:   $10x$