Solve 3t/t+2+5t/t-2-40/t^2-4

$$\frac { 3 t } { t + 2 } + \frac { 5 t } { t - 2 } - \frac { 40 } { t ^ { 2 } - 4 }$$

Evaluate

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

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of t+2 and t-2 is \left(t-2\right)\left(t+2\right). Multiply \frac{3t}{t+2} times \frac{t-2}{t-2}. Multiply \frac{5t}{t-2} times \frac{t+2}{t+2}.

$$\frac{3t\left(t-2\right)}{\left(t-2\right)\left(t+2\right)}+\frac{5t\left(t+2\right)}{\left(t-2\right)\left(t+2\right)}-\frac{40}{t^{2}-4}$$

Since \frac{3t\left(t-2\right)}{\left(t-2\right)\left(t+2\right)} and \frac{5t\left(t+2\right)}{\left(t-2\right)\left(t+2\right)} have the same denominator, add them by adding their numerators.

$$\frac{3t\left(t-2\right)+5t\left(t+2\right)}{\left(t-2\right)\left(t+2\right)}-\frac{40}{t^{2}-4}$$

Do the multiplications in 3t\left(t-2\right)+5t\left(t+2\right).

$$\frac{3t^{2}-6t+5t^{2}+10t}{\left(t-2\right)\left(t+2\right)}-\frac{40}{t^{2}-4}$$

Combine like terms in 3t^{2}-6t+5t^{2}+10t.

$$\frac{8t^{2}+4t}{\left(t-2\right)\left(t+2\right)}-\frac{40}{t^{2}-4}$$

Factor t^{2}-4.

$$\frac{8t^{2}+4t}{\left(t-2\right)\left(t+2\right)}-\frac{40}{\left(t-2\right)\left(t+2\right)}$$

Since \frac{8t^{2}+4t}{\left(t-2\right)\left(t+2\right)} and \frac{40}{\left(t-2\right)\left(t+2\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{8t^{2}+4t-40}{\left(t-2\right)\left(t+2\right)}$$

Factor the expressions that are not already factored in \frac{8t^{2}+4t-40}{\left(t-2\right)\left(t+2\right)}.

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

Cancel out t-2 in both numerator and denominator.

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

Use the distributive property to multiply 4 by 2t+5.

$$\frac{8t+20}{t+2}$$

Differentiate w.r.t. t

$$-\frac{4}{\left(t+2\right)^{2}}$$

Steps Using Derivative Rule for Quotient

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of t+2 and t-2 is \left(t-2\right)\left(t+2\right). Multiply \frac{3t}{t+2} times \frac{t-2}{t-2}. Multiply \frac{5t}{t-2} times \frac{t+2}{t+2}.

$$\frac{\mathrm{d}}{\mathrm{d}t}(\frac{3t\left(t-2\right)}{\left(t-2\right)\left(t+2\right)}+\frac{5t\left(t+2\right)}{\left(t-2\right)\left(t+2\right)}-\frac{40}{t^{2}-4})$$

Since \frac{3t\left(t-2\right)}{\left(t-2\right)\left(t+2\right)} and \frac{5t\left(t+2\right)}{\left(t-2\right)\left(t+2\right)} have the same denominator, add them by adding their numerators.

$$\frac{\mathrm{d}}{\mathrm{d}t}(\frac{3t\left(t-2\right)+5t\left(t+2\right)}{\left(t-2\right)\left(t+2\right)}-\frac{40}{t^{2}-4})$$

Do the multiplications in 3t\left(t-2\right)+5t\left(t+2\right).

$$\frac{\mathrm{d}}{\mathrm{d}t}(\frac{3t^{2}-6t+5t^{2}+10t}{\left(t-2\right)\left(t+2\right)}-\frac{40}{t^{2}-4})$$

Combine like terms in 3t^{2}-6t+5t^{2}+10t.

$$\frac{\mathrm{d}}{\mathrm{d}t}(\frac{8t^{2}+4t}{\left(t-2\right)\left(t+2\right)}-\frac{40}{t^{2}-4})$$

Factor t^{2}-4.

$$\frac{\mathrm{d}}{\mathrm{d}t}(\frac{8t^{2}+4t}{\left(t-2\right)\left(t+2\right)}-\frac{40}{\left(t-2\right)\left(t+2\right)})$$

Since \frac{8t^{2}+4t}{\left(t-2\right)\left(t+2\right)} and \frac{40}{\left(t-2\right)\left(t+2\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\mathrm{d}}{\mathrm{d}t}(\frac{8t^{2}+4t-40}{\left(t-2\right)\left(t+2\right)})$$

Factor the expressions that are not already factored in \frac{8t^{2}+4t-40}{\left(t-2\right)\left(t+2\right)}.

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

Cancel out t-2 in both numerator and denominator.

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

Use the distributive property to multiply 4 by 2t+5.

$$\frac{\mathrm{d}}{\mathrm{d}t}(\frac{8t+20}{t+2})$$

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.

$$\frac{\left(t^{1}+2\right)\frac{\mathrm{d}}{\mathrm{d}t}(8t^{1}+20)-\left(8t^{1}+20\right)\frac{\mathrm{d}}{\mathrm{d}t}(t^{1}+2)}{\left(t^{1}+2\right)^{2}}$$

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}.

$$\frac{\left(t^{1}+2\right)\times 8t^{1-1}-\left(8t^{1}+20\right)t^{1-1}}{\left(t^{1}+2\right)^{2}}$$

Do the arithmetic.

$$\frac{\left(t^{1}+2\right)\times 8t^{0}-\left(8t^{1}+20\right)t^{0}}{\left(t^{1}+2\right)^{2}}$$

Expand using distributive property.

$$\frac{t^{1}\times 8t^{0}+2\times 8t^{0}-\left(8t^{1}t^{0}+20t^{0}\right)}{\left(t^{1}+2\right)^{2}}$$

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

$$\frac{8t^{1}+2\times 8t^{0}-\left(8t^{1}+20t^{0}\right)}{\left(t^{1}+2\right)^{2}}$$

Do the arithmetic.

$$\frac{8t^{1}+16t^{0}-\left(8t^{1}+20t^{0}\right)}{\left(t^{1}+2\right)^{2}}$$

Remove unnecessary parentheses.

$$\frac{8t^{1}+16t^{0}-8t^{1}-20t^{0}}{\left(t^{1}+2\right)^{2}}$$

Combine like terms.

$$\frac{\left(8-8\right)t^{1}+\left(16-20\right)t^{0}}{\left(t^{1}+2\right)^{2}}$$

Subtract 8 from 8 and 20 from 16.

$$\frac{-4t^{0}}{\left(t^{1}+2\right)^{2}}$$

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

$$\frac{-4t^{0}}{\left(t+2\right)^{2}}$$

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

$$\frac{-4}{\left(t+2\right)^{2}}$$