Solve 2/6y-2+y/9y^2-1

$$\frac { 2 } { 6 y - 2 } + \frac { y } { 9 y ^ { 2 } - 1 }$$

Evaluate

$$\frac{4y+1}{9y^{2}-1}$$

Short Solution Steps

Factor 6y-2. Factor 9y^{2}-1.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2\left(3y-1\right) and \left(3y-1\right)\left(3y+1\right) is 2\left(3y-1\right)\left(3y+1\right). Multiply \frac{2}{2\left(3y-1\right)} times \frac{3y+1}{3y+1}. Multiply \frac{y}{\left(3y-1\right)\left(3y+1\right)} times \frac{2}{2}.

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

Since \frac{2\left(3y+1\right)}{2\left(3y-1\right)\left(3y+1\right)} and \frac{2y}{2\left(3y-1\right)\left(3y+1\right)} have the same denominator, add them by adding their numerators.

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

Do the multiplications in 2\left(3y+1\right)+2y.

$$\frac{6y+2+2y}{2\left(3y-1\right)\left(3y+1\right)}$$

Combine like terms in 6y+2+2y.

$$\frac{8y+2}{2\left(3y-1\right)\left(3y+1\right)}$$

Factor the expressions that are not already factored in \frac{8y+2}{2\left(3y-1\right)\left(3y+1\right)}.

$$\frac{2\left(4y+1\right)}{2\left(3y-1\right)\left(3y+1\right)}$$

Cancel out 2 in both numerator and denominator.

$$\frac{4y+1}{\left(3y-1\right)\left(3y+1\right)}$$

Expand \left(3y-1\right)\left(3y+1\right).

$$\frac{4y+1}{9y^{2}-1}$$

Differentiate w.r.t. y

$$-\frac{2\left(18y^{2}+9y+2\right)}{\left(9y^{2}-1\right)^{2}}$$

Steps Using Derivative Rule for Quotient

Factor 6y-2. Factor 9y^{2}-1.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2\left(3y-1\right) and \left(3y-1\right)\left(3y+1\right) is 2\left(3y-1\right)\left(3y+1\right). Multiply \frac{2}{2\left(3y-1\right)} times \frac{3y+1}{3y+1}. Multiply \frac{y}{\left(3y-1\right)\left(3y+1\right)} times \frac{2}{2}.

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

Since \frac{2\left(3y+1\right)}{2\left(3y-1\right)\left(3y+1\right)} and \frac{2y}{2\left(3y-1\right)\left(3y+1\right)} have the same denominator, add them by adding their numerators.

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

Do the multiplications in 2\left(3y+1\right)+2y.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{6y+2+2y}{2\left(3y-1\right)\left(3y+1\right)})$$

Combine like terms in 6y+2+2y.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{8y+2}{2\left(3y-1\right)\left(3y+1\right)})$$

Factor the expressions that are not already factored in \frac{8y+2}{2\left(3y-1\right)\left(3y+1\right)}.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{2\left(4y+1\right)}{2\left(3y-1\right)\left(3y+1\right)})$$

Cancel out 2 in both numerator and denominator.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{4y+1}{\left(3y-1\right)\left(3y+1\right)})$$

Consider \left(3y-1\right)\left(3y+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{4y+1}{\left(3y\right)^{2}-1})$$

Expand \left(3y\right)^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{4y+1}{3^{2}y^{2}-1})$$

Calculate 3 to the power of 2 and get 9.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{4y+1}{9y^{2}-1})$$

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.

$$\frac{\left(9y^{2}-1\right)\frac{\mathrm{d}}{\mathrm{d}y}(4y^{1}+1)-\left(4y^{1}+1\right)\frac{\mathrm{d}}{\mathrm{d}y}(9y^{2}-1)}{\left(9y^{2}-1\right)^{2}}$$

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}.

$$\frac{\left(9y^{2}-1\right)\times 4y^{1-1}-\left(4y^{1}+1\right)\times 2\times 9y^{2-1}}{\left(9y^{2}-1\right)^{2}}$$

Do the arithmetic.

$$\frac{\left(9y^{2}-1\right)\times 4y^{0}-\left(4y^{1}+1\right)\times 18y^{1}}{\left(9y^{2}-1\right)^{2}}$$

Expand using distributive property.

$$\frac{9y^{2}\times 4y^{0}-4y^{0}-\left(4y^{1}\times 18y^{1}+18y^{1}\right)}{\left(9y^{2}-1\right)^{2}}$$

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

$$\frac{9\times 4y^{2}-4y^{0}-\left(4\times 18y^{1+1}+18y^{1}\right)}{\left(9y^{2}-1\right)^{2}}$$

Do the arithmetic.

$$\frac{36y^{2}-4y^{0}-\left(72y^{2}+18y^{1}\right)}{\left(9y^{2}-1\right)^{2}}$$

Remove unnecessary parentheses.

$$\frac{36y^{2}-4y^{0}-72y^{2}-18y^{1}}{\left(9y^{2}-1\right)^{2}}$$

Combine like terms.

$$\frac{\left(36-72\right)y^{2}-4y^{0}-18y^{1}}{\left(9y^{2}-1\right)^{2}}$$

Subtract 72 from 36.

$$\frac{-36y^{2}-4y^{0}-18y^{1}}{\left(9y^{2}-1\right)^{2}}$$

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

$$\frac{-36y^{2}-4y^{0}-18y}{\left(9y^{2}-1\right)^{2}}$$

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

$$\frac{-36y^{2}-4-18y}{\left(9y^{2}-1\right)^{2}}$$