⚠ yx is read as the product y*x: for the variable yx, write yx_■(\partial)/(\partial x)(2yx-3y+y^2) is read as dif(2yx-3y+y^2,x): use dif(expression, variable) for a derivative with respect to that variable, use (numerator)/(denominator) for a quotient

Differentiate with respect to x while holding y constant

Let's differentiate: $\frac{\mathrm{\partial }}{\mathrm{\partial }x}(2yx-3y+y^2)$

ⓘ The derivative of a sum is the sum of the derivatives: $\frac{\mathrm{\partial }}{\mathrm{\partial }x}(2yx-3y+y^2)$ = $\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(2yx\right)+\frac{\mathrm{\partial }}{\mathrm{\partial }x}(-3y)+\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(y^2\right)$

$\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(2yx\right)+\frac{\mathrm{\partial }}{\mathrm{\partial }x}(-3y)+\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(y^2\right)$

ⓘ For differentiation with respect to $x$, $2y$ is treated as constant. A constant factor stays in front of the derivative, (k*u)' = k*u': $\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(2yx\right)$ = $2y·\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(x\right)$■The derivative of a constant is zero: $\frac{\mathrm{\partial }}{\mathrm{\partial }x}(-3y)$ = $0$, $\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(y^2\right)$ = $0$

$2y·\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(x\right)+0+0$

ⓘ Adding 0 changes nothing: $2y·\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(x\right)+0+0$ = $2y·\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(x\right)$■The derivative of x is 1: $\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(x\right)$ = $1$

$2y·1$

▶Derivative:   $2y$