Solve (x+y-x/1+xy):(1+x^2-xy/1+xy)

$$( x + \frac { y - x } { 1 + x y } ) : ( 1 + \frac { x ^ { 2 } - x y } { 1 + x y } )$$

Evaluate

$$y$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{1+xy}{1+xy}.

$$\frac{\frac{x\left(1+xy\right)}{1+xy}+\frac{y-x}{1+xy}}{1+\frac{x^{2}-xy}{1+xy}}$$

Since \frac{x\left(1+xy\right)}{1+xy} and \frac{y-x}{1+xy} have the same denominator, add them by adding their numerators.

$$\frac{\frac{x\left(1+xy\right)+y-x}{1+xy}}{1+\frac{x^{2}-xy}{1+xy}}$$

Do the multiplications in x\left(1+xy\right)+y-x.

$$\frac{\frac{x+x^{2}y+y-x}{1+xy}}{1+\frac{x^{2}-xy}{1+xy}}$$

Combine like terms in x+x^{2}y+y-x.

$$\frac{\frac{y+x^{2}y}{1+xy}}{1+\frac{x^{2}-xy}{1+xy}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{1+xy}{1+xy}.

$$\frac{\frac{y+x^{2}y}{1+xy}}{\frac{1+xy}{1+xy}+\frac{x^{2}-xy}{1+xy}}$$

Since \frac{1+xy}{1+xy} and \frac{x^{2}-xy}{1+xy} have the same denominator, add them by adding their numerators.

$$\frac{\frac{y+x^{2}y}{1+xy}}{\frac{1+xy+x^{2}-xy}{1+xy}}$$

Combine like terms in 1+xy+x^{2}-xy.

$$\frac{\frac{y+x^{2}y}{1+xy}}{\frac{1+x^{2}}{1+xy}}$$

Divide \frac{y+x^{2}y}{1+xy} by \frac{1+x^{2}}{1+xy} by multiplying \frac{y+x^{2}y}{1+xy} by the reciprocal of \frac{1+x^{2}}{1+xy}.

$$\frac{\left(y+x^{2}y\right)\left(1+xy\right)}{\left(1+xy\right)\left(1+x^{2}\right)}$$

Cancel out xy+1 in both numerator and denominator.

$$\frac{yx^{2}+y}{x^{2}+1}$$

Factor the expressions that are not already factored.

$$\frac{y\left(x^{2}+1\right)}{x^{2}+1}$$

Cancel out x^{2}+1 in both numerator and denominator.

$$y$$

Differentiate w.r.t. y

$$1$$

Steps Using Definition of a Derivative

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{1+xy}{1+xy}.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{\frac{x\left(1+xy\right)}{1+xy}+\frac{y-x}{1+xy}}{1+\frac{x^{2}-xy}{1+xy}})$$

Since \frac{x\left(1+xy\right)}{1+xy} and \frac{y-x}{1+xy} have the same denominator, add them by adding their numerators.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{\frac{x\left(1+xy\right)+y-x}{1+xy}}{1+\frac{x^{2}-xy}{1+xy}})$$

Do the multiplications in x\left(1+xy\right)+y-x.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{\frac{x+x^{2}y+y-x}{1+xy}}{1+\frac{x^{2}-xy}{1+xy}})$$

Combine like terms in x+x^{2}y+y-x.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{\frac{y+x^{2}y}{1+xy}}{1+\frac{x^{2}-xy}{1+xy}})$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{1+xy}{1+xy}.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{\frac{y+x^{2}y}{1+xy}}{\frac{1+xy}{1+xy}+\frac{x^{2}-xy}{1+xy}})$$

Since \frac{1+xy}{1+xy} and \frac{x^{2}-xy}{1+xy} have the same denominator, add them by adding their numerators.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{\frac{y+x^{2}y}{1+xy}}{\frac{1+xy+x^{2}-xy}{1+xy}})$$

Combine like terms in 1+xy+x^{2}-xy.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{\frac{y+x^{2}y}{1+xy}}{\frac{1+x^{2}}{1+xy}})$$

Divide \frac{y+x^{2}y}{1+xy} by \frac{1+x^{2}}{1+xy} by multiplying \frac{y+x^{2}y}{1+xy} by the reciprocal of \frac{1+x^{2}}{1+xy}.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{\left(y+x^{2}y\right)\left(1+xy\right)}{\left(1+xy\right)\left(1+x^{2}\right)})$$

Cancel out xy+1 in both numerator and denominator.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{yx^{2}+y}{x^{2}+1})$$

Factor the expressions that are not already factored in \frac{yx^{2}+y}{x^{2}+1}.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{y\left(x^{2}+1\right)}{x^{2}+1})$$

Cancel out x^{2}+1 in both numerator and denominator.

$$\frac{\mathrm{d}}{\mathrm{d}y}(y)$$

The derivative of ax^{n} is nax^{n-1}.

$$y^{1-1}$$

Subtract 1 from 1.

$$y^{0}$$

For any term t except 0, t^{0}=1.

$$1$$