Solve 5/y+5-1/y+2

$$\frac{ 5 }{ y+5 } - \frac{ 1 }{ y+2 }$$

Evaluate

$$\frac{4y+5}{\left(y+2\right)\left(y+5\right)}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of y+5 and y+2 is \left(y+2\right)\left(y+5\right). Multiply \frac{5}{y+5} times \frac{y+2}{y+2}. Multiply \frac{1}{y+2} times \frac{y+5}{y+5}.

$$\frac{5\left(y+2\right)}{\left(y+2\right)\left(y+5\right)}-\frac{y+5}{\left(y+2\right)\left(y+5\right)}$$

Since \frac{5\left(y+2\right)}{\left(y+2\right)\left(y+5\right)} and \frac{y+5}{\left(y+2\right)\left(y+5\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{5\left(y+2\right)-\left(y+5\right)}{\left(y+2\right)\left(y+5\right)}$$

Do the multiplications in 5\left(y+2\right)-\left(y+5\right).

$$\frac{5y+10-y-5}{\left(y+2\right)\left(y+5\right)}$$

Combine like terms in 5y+10-y-5.

$$\frac{4y+5}{\left(y+2\right)\left(y+5\right)}$$

Expand \left(y+2\right)\left(y+5\right).

$$\frac{4y+5}{y^{2}+7y+10}$$

Differentiate w.r.t. y

$$\frac{5-10y-4y^{2}}{y^{4}+14y^{3}+69y^{2}+140y+100}$$

Steps Using Derivative Rule for Quotient

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of y+5 and y+2 is \left(y+2\right)\left(y+5\right). Multiply \frac{5}{y+5} times \frac{y+2}{y+2}. Multiply \frac{1}{y+2} times \frac{y+5}{y+5}.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{5\left(y+2\right)}{\left(y+2\right)\left(y+5\right)}-\frac{y+5}{\left(y+2\right)\left(y+5\right)})$$

Since \frac{5\left(y+2\right)}{\left(y+2\right)\left(y+5\right)} and \frac{y+5}{\left(y+2\right)\left(y+5\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{5\left(y+2\right)-\left(y+5\right)}{\left(y+2\right)\left(y+5\right)})$$

Do the multiplications in 5\left(y+2\right)-\left(y+5\right).

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{5y+10-y-5}{\left(y+2\right)\left(y+5\right)})$$

Combine like terms in 5y+10-y-5.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{4y+5}{\left(y+2\right)\left(y+5\right)})$$

Apply the distributive property by multiplying each term of y+2 by each term of y+5.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{4y+5}{y^{2}+5y+2y+10})$$

Combine 5y and 2y to get 7y.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{4y+5}{y^{2}+7y+10})$$

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.

$$\frac{\left(y^{2}+7y^{1}+10\right)\frac{\mathrm{d}}{\mathrm{d}y}(4y^{1}+5)-\left(4y^{1}+5\right)\frac{\mathrm{d}}{\mathrm{d}y}(y^{2}+7y^{1}+10)}{\left(y^{2}+7y^{1}+10\right)^{2}}$$

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}.

$$\frac{\left(y^{2}+7y^{1}+10\right)\times 4y^{1-1}-\left(4y^{1}+5\right)\left(2y^{2-1}+7y^{1-1}\right)}{\left(y^{2}+7y^{1}+10\right)^{2}}$$

Simplify.

$$\frac{\left(y^{2}+7y^{1}+10\right)\times 4y^{0}-\left(4y^{1}+5\right)\left(2y^{1}+7y^{0}\right)}{\left(y^{2}+7y^{1}+10\right)^{2}}$$

Multiply y^{2}+7y^{1}+10 times 4y^{0}.

$$\frac{y^{2}\times 4y^{0}+7y^{1}\times 4y^{0}+10\times 4y^{0}-\left(4y^{1}+5\right)\left(2y^{1}+7y^{0}\right)}{\left(y^{2}+7y^{1}+10\right)^{2}}$$

Multiply 4y^{1}+5 times 2y^{1}+7y^{0}.

$$\frac{y^{2}\times 4y^{0}+7y^{1}\times 4y^{0}+10\times 4y^{0}-\left(4y^{1}\times 2y^{1}+4y^{1}\times 7y^{0}+5\times 2y^{1}+5\times 7y^{0}\right)}{\left(y^{2}+7y^{1}+10\right)^{2}}$$

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

$$\frac{4y^{2}+7\times 4y^{1}+10\times 4y^{0}-\left(4\times 2y^{1+1}+4\times 7y^{1}+5\times 2y^{1}+5\times 7y^{0}\right)}{\left(y^{2}+7y^{1}+10\right)^{2}}$$

Simplify.

$$\frac{4y^{2}+28y^{1}+40y^{0}-\left(8y^{2}+28y^{1}+10y^{1}+35y^{0}\right)}{\left(y^{2}+7y^{1}+10\right)^{2}}$$

Combine like terms.

$$\frac{-4y^{2}-10y^{1}+5y^{0}}{\left(y^{2}+7y^{1}+10\right)^{2}}$$

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

$$\frac{-4y^{2}-10y+5y^{0}}{\left(y^{2}+7y+10\right)^{2}}$$

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

$$\frac{-4y^{2}-10y+5\times 1}{\left(y^{2}+7y+10\right)^{2}}$$

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

$$\frac{-4y^{2}-10y+5}{\left(y^{2}+7y+10\right)^{2}}$$