Solve 3xy/xy+x

$$\frac { 3 x y } { x y + x }$$

Evaluate

$$\frac{3y}{y+1}$$

Short Solution Steps

Factor the expressions that are not already factored.

$$\frac{3xy}{x\left(y+1\right)}$$

Cancel out x in both numerator and denominator.

$$\frac{3y}{y+1}$$

Differentiate w.r.t. y

$$\frac{3}{\left(y+1\right)^{2}}$$

Steps Using Derivative Rule for Quotient

Factor the expressions that are not already factored in \frac{3xy}{xy+x}.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{3xy}{x\left(y+1\right)})$$

Cancel out x in both numerator and denominator.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{3y}{y+1})$$

For any two differentiable functions, the derivative of the quotient of two functions is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the denominator squared.

$$\frac{\left(y^{1}+1\right)\frac{\mathrm{d}}{\mathrm{d}y}(3y^{1})-3y^{1}\frac{\mathrm{d}}{\mathrm{d}y}(y^{1}+1)}{\left(y^{1}+1\right)^{2}}$$

The derivative of a polynomial is the sum of the derivatives of its terms. The derivative of a constant term is 0. The derivative of ax^{n} is nax^{n-1}.

$$\frac{\left(y^{1}+1\right)\times 3y^{1-1}-3y^{1}y^{1-1}}{\left(y^{1}+1\right)^{2}}$$

Do the arithmetic.

$$\frac{\left(y^{1}+1\right)\times 3y^{0}-3y^{1}y^{0}}{\left(y^{1}+1\right)^{2}}$$

Expand using distributive property.

$$\frac{y^{1}\times 3y^{0}+3y^{0}-3y^{1}y^{0}}{\left(y^{1}+1\right)^{2}}$$

To multiply powers of the same base, add their exponents.

$$\frac{3y^{1}+3y^{0}-3y^{1}}{\left(y^{1}+1\right)^{2}}$$

Combine like terms.

$$\frac{\left(3-3\right)y^{1}+3y^{0}}{\left(y^{1}+1\right)^{2}}$$

Subtract 3 from 3.

$$\frac{3y^{0}}{\left(y^{1}+1\right)^{2}}$$

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

$$\frac{3y^{0}}{\left(y+1\right)^{2}}$$

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

$$\frac{3\times 1}{\left(y+1\right)^{2}}$$

For any term t, t\times 1=t and 1t=t.

$$\frac{3}{\left(y+1\right)^{2}}$$