Solve 2/y^2-16-3/y+4

$$\frac { 2 } { y ^ { 2 } - 16 } - \frac { 3 } { y + 4 }$$

Evaluate

$$\frac{14-3y}{y^{2}-16}$$

Short Solution Steps

Factor y^{2}-16.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(y-4\right)\left(y+4\right) and y+4 is \left(y-4\right)\left(y+4\right). Multiply \frac{3}{y+4} times \frac{y-4}{y-4}.

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

Since \frac{2}{\left(y-4\right)\left(y+4\right)} and \frac{3\left(y-4\right)}{\left(y-4\right)\left(y+4\right)} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 2-3\left(y-4\right).

$$\frac{2-3y+12}{\left(y-4\right)\left(y+4\right)}$$

Combine like terms in 2-3y+12.

$$\frac{14-3y}{\left(y-4\right)\left(y+4\right)}$$

Expand \left(y-4\right)\left(y+4\right).

$$\frac{14-3y}{y^{2}-16}$$

Differentiate w.r.t. y

$$\frac{3y^{2}-28y+48}{\left(y^{2}-16\right)^{2}}$$

Steps Using Derivative Rule for Quotient

Factor y^{2}-16.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(y-4\right)\left(y+4\right) and y+4 is \left(y-4\right)\left(y+4\right). Multiply \frac{3}{y+4} times \frac{y-4}{y-4}.

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

Since \frac{2}{\left(y-4\right)\left(y+4\right)} and \frac{3\left(y-4\right)}{\left(y-4\right)\left(y+4\right)} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 2-3\left(y-4\right).

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{2-3y+12}{\left(y-4\right)\left(y+4\right)})$$

Combine like terms in 2-3y+12.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{14-3y}{\left(y-4\right)\left(y+4\right)})$$

Consider \left(y-4\right)\left(y+4\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 4.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{14-3y}{y^{2}-16})$$

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.

$$\frac{\left(y^{2}-16\right)\frac{\mathrm{d}}{\mathrm{d}y}(-3y^{1}+14)-\left(-3y^{1}+14\right)\frac{\mathrm{d}}{\mathrm{d}y}(y^{2}-16)}{\left(y^{2}-16\right)^{2}}$$

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}.

$$\frac{\left(y^{2}-16\right)\left(-3\right)y^{1-1}-\left(-3y^{1}+14\right)\times 2y^{2-1}}{\left(y^{2}-16\right)^{2}}$$

Do the arithmetic.

$$\frac{\left(y^{2}-16\right)\left(-3\right)y^{0}-\left(-3y^{1}+14\right)\times 2y^{1}}{\left(y^{2}-16\right)^{2}}$$

Expand using distributive property.

$$\frac{y^{2}\left(-3\right)y^{0}-16\left(-3\right)y^{0}-\left(-3y^{1}\times 2y^{1}+14\times 2y^{1}\right)}{\left(y^{2}-16\right)^{2}}$$

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

$$\frac{-3y^{2}-16\left(-3\right)y^{0}-\left(-3\times 2y^{1+1}+14\times 2y^{1}\right)}{\left(y^{2}-16\right)^{2}}$$

Do the arithmetic.

$$\frac{-3y^{2}+48y^{0}-\left(-6y^{2}+28y^{1}\right)}{\left(y^{2}-16\right)^{2}}$$

Remove unnecessary parentheses.

$$\frac{-3y^{2}+48y^{0}-\left(-6y^{2}\right)-28y^{1}}{\left(y^{2}-16\right)^{2}}$$

Combine like terms.

$$\frac{\left(-3-\left(-6\right)\right)y^{2}+48y^{0}-28y^{1}}{\left(y^{2}-16\right)^{2}}$$

Subtract -6 from -3.

$$\frac{3y^{2}+48y^{0}-28y^{1}}{\left(y^{2}-16\right)^{2}}$$

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

$$\frac{3y^{2}+48y^{0}-28y}{\left(y^{2}-16\right)^{2}}$$

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

$$\frac{3y^{2}+48\times 1-28y}{\left(y^{2}-16\right)^{2}}$$

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

$$\frac{3y^{2}+48-28y}{\left(y^{2}-16\right)^{2}}$$