Solve 8/d-3/frac{10{d+3}+6/d^2-9}

$$\frac { \frac { 8 } { d - 3 } } { \frac { 10 } { d + 3 } + \frac { 6 } { d ^ { 2 } - 9 } }$$

Evaluate

$$\frac{4\left(d+3\right)}{5d-12}$$

Short Solution Steps

Factor d^{2}-9.

$$\frac{\frac{8}{d-3}}{\frac{10}{d+3}+\frac{6}{\left(d-3\right)\left(d+3\right)}}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of d+3 and \left(d-3\right)\left(d+3\right) is \left(d-3\right)\left(d+3\right). Multiply \frac{10}{d+3} times \frac{d-3}{d-3}.

$$\frac{\frac{8}{d-3}}{\frac{10\left(d-3\right)}{\left(d-3\right)\left(d+3\right)}+\frac{6}{\left(d-3\right)\left(d+3\right)}}$$

Since \frac{10\left(d-3\right)}{\left(d-3\right)\left(d+3\right)} and \frac{6}{\left(d-3\right)\left(d+3\right)} have the same denominator, add them by adding their numerators.

$$\frac{\frac{8}{d-3}}{\frac{10\left(d-3\right)+6}{\left(d-3\right)\left(d+3\right)}}$$

Do the multiplications in 10\left(d-3\right)+6.

$$\frac{\frac{8}{d-3}}{\frac{10d-30+6}{\left(d-3\right)\left(d+3\right)}}$$

Combine like terms in 10d-30+6.

$$\frac{\frac{8}{d-3}}{\frac{10d-24}{\left(d-3\right)\left(d+3\right)}}$$

Divide \frac{8}{d-3} by \frac{10d-24}{\left(d-3\right)\left(d+3\right)} by multiplying \frac{8}{d-3} by the reciprocal of \frac{10d-24}{\left(d-3\right)\left(d+3\right)}.

$$\frac{8\left(d-3\right)\left(d+3\right)}{\left(d-3\right)\left(10d-24\right)}$$

Cancel out d-3 in both numerator and denominator.

$$\frac{8\left(d+3\right)}{10d-24}$$

Factor the expressions that are not already factored.

$$\frac{8\left(d+3\right)}{2\left(5d-12\right)}$$

Cancel out 2 in both numerator and denominator.

$$\frac{4\left(d+3\right)}{5d-12}$$

Expand the expression.

$$\frac{4d+12}{5d-12}$$

Differentiate w.r.t. d

$$-\frac{108}{\left(5d-12\right)^{2}}$$

Steps Using Derivative Rule for Quotient

Factor d^{2}-9.

$$\frac{\mathrm{d}}{\mathrm{d}d}(\frac{\frac{8}{d-3}}{\frac{10}{d+3}+\frac{6}{\left(d-3\right)\left(d+3\right)}})$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of d+3 and \left(d-3\right)\left(d+3\right) is \left(d-3\right)\left(d+3\right). Multiply \frac{10}{d+3} times \frac{d-3}{d-3}.

$$\frac{\mathrm{d}}{\mathrm{d}d}(\frac{\frac{8}{d-3}}{\frac{10\left(d-3\right)}{\left(d-3\right)\left(d+3\right)}+\frac{6}{\left(d-3\right)\left(d+3\right)}})$$

Since \frac{10\left(d-3\right)}{\left(d-3\right)\left(d+3\right)} and \frac{6}{\left(d-3\right)\left(d+3\right)} have the same denominator, add them by adding their numerators.

$$\frac{\mathrm{d}}{\mathrm{d}d}(\frac{\frac{8}{d-3}}{\frac{10\left(d-3\right)+6}{\left(d-3\right)\left(d+3\right)}})$$

Do the multiplications in 10\left(d-3\right)+6.

$$\frac{\mathrm{d}}{\mathrm{d}d}(\frac{\frac{8}{d-3}}{\frac{10d-30+6}{\left(d-3\right)\left(d+3\right)}})$$

Combine like terms in 10d-30+6.

$$\frac{\mathrm{d}}{\mathrm{d}d}(\frac{\frac{8}{d-3}}{\frac{10d-24}{\left(d-3\right)\left(d+3\right)}})$$

Divide \frac{8}{d-3} by \frac{10d-24}{\left(d-3\right)\left(d+3\right)} by multiplying \frac{8}{d-3} by the reciprocal of \frac{10d-24}{\left(d-3\right)\left(d+3\right)}.

$$\frac{\mathrm{d}}{\mathrm{d}d}(\frac{8\left(d-3\right)\left(d+3\right)}{\left(d-3\right)\left(10d-24\right)})$$

Cancel out d-3 in both numerator and denominator.

$$\frac{\mathrm{d}}{\mathrm{d}d}(\frac{8\left(d+3\right)}{10d-24})$$

Factor the expressions that are not already factored in \frac{8\left(d+3\right)}{10d-24}.

$$\frac{\mathrm{d}}{\mathrm{d}d}(\frac{8\left(d+3\right)}{2\left(5d-12\right)})$$

Cancel out 2 in both numerator and denominator.

$$\frac{\mathrm{d}}{\mathrm{d}d}(\frac{4\left(d+3\right)}{5d-12})$$

Use the distributive property to multiply 4 by d+3.

$$\frac{\mathrm{d}}{\mathrm{d}d}(\frac{4d+12}{5d-12})$$

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.

$$\frac{\left(5d^{1}-12\right)\frac{\mathrm{d}}{\mathrm{d}d}(4d^{1}+12)-\left(4d^{1}+12\right)\frac{\mathrm{d}}{\mathrm{d}d}(5d^{1}-12)}{\left(5d^{1}-12\right)^{2}}$$

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}.

$$\frac{\left(5d^{1}-12\right)\times 4d^{1-1}-\left(4d^{1}+12\right)\times 5d^{1-1}}{\left(5d^{1}-12\right)^{2}}$$

Do the arithmetic.

$$\frac{\left(5d^{1}-12\right)\times 4d^{0}-\left(4d^{1}+12\right)\times 5d^{0}}{\left(5d^{1}-12\right)^{2}}$$

Expand using distributive property.

$$\frac{5d^{1}\times 4d^{0}-12\times 4d^{0}-\left(4d^{1}\times 5d^{0}+12\times 5d^{0}\right)}{\left(5d^{1}-12\right)^{2}}$$

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

$$\frac{5\times 4d^{1}-12\times 4d^{0}-\left(4\times 5d^{1}+12\times 5d^{0}\right)}{\left(5d^{1}-12\right)^{2}}$$

Do the arithmetic.

$$\frac{20d^{1}-48d^{0}-\left(20d^{1}+60d^{0}\right)}{\left(5d^{1}-12\right)^{2}}$$

Remove unnecessary parentheses.

$$\frac{20d^{1}-48d^{0}-20d^{1}-60d^{0}}{\left(5d^{1}-12\right)^{2}}$$

Combine like terms.

$$\frac{\left(20-20\right)d^{1}+\left(-48-60\right)d^{0}}{\left(5d^{1}-12\right)^{2}}$$

Subtract 20 from 20 and 60 from -48.

$$\frac{-108d^{0}}{\left(5d^{1}-12\right)^{2}}$$

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

$$\frac{-108d^{0}}{\left(5d-12\right)^{2}}$$

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

$$\frac{-108}{\left(5d-12\right)^{2}}$$