Solve t/t+3+4t/t-3-18/t^2-9

$$\frac { t } { t + 3 } + \frac { 4 t } { t - 3 } - \frac { 18 } { t ^ { 2 } - 9 }$$

Evaluate

$$\frac{5t-6}{t-3}$$

Short Solution Steps

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

$$\frac{t\left(t-3\right)}{\left(t-3\right)\left(t+3\right)}+\frac{4t\left(t+3\right)}{\left(t-3\right)\left(t+3\right)}-\frac{18}{t^{2}-9}$$

Since \frac{t\left(t-3\right)}{\left(t-3\right)\left(t+3\right)} and \frac{4t\left(t+3\right)}{\left(t-3\right)\left(t+3\right)} have the same denominator, add them by adding their numerators.

$$\frac{t\left(t-3\right)+4t\left(t+3\right)}{\left(t-3\right)\left(t+3\right)}-\frac{18}{t^{2}-9}$$

Do the multiplications in t\left(t-3\right)+4t\left(t+3\right).

$$\frac{t^{2}-3t+4t^{2}+12t}{\left(t-3\right)\left(t+3\right)}-\frac{18}{t^{2}-9}$$

Combine like terms in t^{2}-3t+4t^{2}+12t.

$$\frac{5t^{2}+9t}{\left(t-3\right)\left(t+3\right)}-\frac{18}{t^{2}-9}$$

Factor t^{2}-9.

$$\frac{5t^{2}+9t}{\left(t-3\right)\left(t+3\right)}-\frac{18}{\left(t-3\right)\left(t+3\right)}$$

Since \frac{5t^{2}+9t}{\left(t-3\right)\left(t+3\right)} and \frac{18}{\left(t-3\right)\left(t+3\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{5t^{2}+9t-18}{\left(t-3\right)\left(t+3\right)}$$

Factor the expressions that are not already factored in \frac{5t^{2}+9t-18}{\left(t-3\right)\left(t+3\right)}.

$$\frac{\left(5t-6\right)\left(t+3\right)}{\left(t-3\right)\left(t+3\right)}$$

Cancel out t+3 in both numerator and denominator.

$$\frac{5t-6}{t-3}$$

Differentiate w.r.t. t

$$-\frac{9}{\left(t-3\right)^{2}}$$

Steps Using Derivative Rule for Quotient

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

$$\frac{\mathrm{d}}{\mathrm{d}t}(\frac{t\left(t-3\right)}{\left(t-3\right)\left(t+3\right)}+\frac{4t\left(t+3\right)}{\left(t-3\right)\left(t+3\right)}-\frac{18}{t^{2}-9})$$

Since \frac{t\left(t-3\right)}{\left(t-3\right)\left(t+3\right)} and \frac{4t\left(t+3\right)}{\left(t-3\right)\left(t+3\right)} have the same denominator, add them by adding their numerators.

$$\frac{\mathrm{d}}{\mathrm{d}t}(\frac{t\left(t-3\right)+4t\left(t+3\right)}{\left(t-3\right)\left(t+3\right)}-\frac{18}{t^{2}-9})$$

Do the multiplications in t\left(t-3\right)+4t\left(t+3\right).

$$\frac{\mathrm{d}}{\mathrm{d}t}(\frac{t^{2}-3t+4t^{2}+12t}{\left(t-3\right)\left(t+3\right)}-\frac{18}{t^{2}-9})$$

Combine like terms in t^{2}-3t+4t^{2}+12t.

$$\frac{\mathrm{d}}{\mathrm{d}t}(\frac{5t^{2}+9t}{\left(t-3\right)\left(t+3\right)}-\frac{18}{t^{2}-9})$$

Factor t^{2}-9.

$$\frac{\mathrm{d}}{\mathrm{d}t}(\frac{5t^{2}+9t}{\left(t-3\right)\left(t+3\right)}-\frac{18}{\left(t-3\right)\left(t+3\right)})$$

Since \frac{5t^{2}+9t}{\left(t-3\right)\left(t+3\right)} and \frac{18}{\left(t-3\right)\left(t+3\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\mathrm{d}}{\mathrm{d}t}(\frac{5t^{2}+9t-18}{\left(t-3\right)\left(t+3\right)})$$

Factor the expressions that are not already factored in \frac{5t^{2}+9t-18}{\left(t-3\right)\left(t+3\right)}.

$$\frac{\mathrm{d}}{\mathrm{d}t}(\frac{\left(5t-6\right)\left(t+3\right)}{\left(t-3\right)\left(t+3\right)})$$

Cancel out t+3 in both numerator and denominator.

$$\frac{\mathrm{d}}{\mathrm{d}t}(\frac{5t-6}{t-3})$$

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.

$$\frac{\left(t^{1}-3\right)\frac{\mathrm{d}}{\mathrm{d}t}(5t^{1}-6)-\left(5t^{1}-6\right)\frac{\mathrm{d}}{\mathrm{d}t}(t^{1}-3)}{\left(t^{1}-3\right)^{2}}$$

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}.

$$\frac{\left(t^{1}-3\right)\times 5t^{1-1}-\left(5t^{1}-6\right)t^{1-1}}{\left(t^{1}-3\right)^{2}}$$

Do the arithmetic.

$$\frac{\left(t^{1}-3\right)\times 5t^{0}-\left(5t^{1}-6\right)t^{0}}{\left(t^{1}-3\right)^{2}}$$

Expand using distributive property.

$$\frac{t^{1}\times 5t^{0}-3\times 5t^{0}-\left(5t^{1}t^{0}-6t^{0}\right)}{\left(t^{1}-3\right)^{2}}$$

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

$$\frac{5t^{1}-3\times 5t^{0}-\left(5t^{1}-6t^{0}\right)}{\left(t^{1}-3\right)^{2}}$$

Do the arithmetic.

$$\frac{5t^{1}-15t^{0}-\left(5t^{1}-6t^{0}\right)}{\left(t^{1}-3\right)^{2}}$$

Remove unnecessary parentheses.

$$\frac{5t^{1}-15t^{0}-5t^{1}-\left(-6t^{0}\right)}{\left(t^{1}-3\right)^{2}}$$

Combine like terms.

$$\frac{\left(5-5\right)t^{1}+\left(-15-\left(-6\right)\right)t^{0}}{\left(t^{1}-3\right)^{2}}$$

Subtract 5 from 5 and -6 from -15.

$$\frac{-9t^{0}}{\left(t^{1}-3\right)^{2}}$$

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

$$\frac{-9t^{0}}{\left(t-3\right)^{2}}$$

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

$$\frac{-9}{\left(t-3\right)^{2}}$$