Solve 6/2k+4+2/k-3

$$\frac { 6 } { 2 k + 4 } + \frac { 2 } { k - 3 }$$

Evaluate

$$\frac{5\left(k-1\right)}{\left(k-3\right)\left(k+2\right)}$$

Short Solution Steps

Factor 2k+4.

$$\frac{6}{2\left(k+2\right)}+\frac{2}{k-3}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2\left(k+2\right) and k-3 is 2\left(k-3\right)\left(k+2\right). Multiply \frac{6}{2\left(k+2\right)} times \frac{k-3}{k-3}. Multiply \frac{2}{k-3} times \frac{2\left(k+2\right)}{2\left(k+2\right)}.

$$\frac{6\left(k-3\right)}{2\left(k-3\right)\left(k+2\right)}+\frac{2\times 2\left(k+2\right)}{2\left(k-3\right)\left(k+2\right)}$$

Since \frac{6\left(k-3\right)}{2\left(k-3\right)\left(k+2\right)} and \frac{2\times 2\left(k+2\right)}{2\left(k-3\right)\left(k+2\right)} have the same denominator, add them by adding their numerators.

$$\frac{6\left(k-3\right)+2\times 2\left(k+2\right)}{2\left(k-3\right)\left(k+2\right)}$$

Do the multiplications in 6\left(k-3\right)+2\times 2\left(k+2\right).

$$\frac{6k-18+4k+8}{2\left(k-3\right)\left(k+2\right)}$$

Combine like terms in 6k-18+4k+8.

$$\frac{10k-10}{2\left(k-3\right)\left(k+2\right)}$$

Factor the expressions that are not already factored in \frac{10k-10}{2\left(k-3\right)\left(k+2\right)}.

$$\frac{10\left(k-1\right)}{2\left(k-3\right)\left(k+2\right)}$$

Cancel out 2 in both numerator and denominator.

$$\frac{5\left(k-1\right)}{\left(k-3\right)\left(k+2\right)}$$

Expand \left(k-3\right)\left(k+2\right).

$$\frac{5\left(k-1\right)}{k^{2}-k-6}$$

Use the distributive property to multiply 5 by k-1.

$$\frac{5k-5}{k^{2}-k-6}$$

Differentiate w.r.t. k

$$\frac{5\left(-k^{2}+2k-7\right)}{\left(\left(k-3\right)\left(k+2\right)\right)^{2}}$$

Steps Using Derivative Rule for Quotient

Factor 2k+4.

$$\frac{\mathrm{d}}{\mathrm{d}k}(\frac{6}{2\left(k+2\right)}+\frac{2}{k-3})$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2\left(k+2\right) and k-3 is 2\left(k-3\right)\left(k+2\right). Multiply \frac{6}{2\left(k+2\right)} times \frac{k-3}{k-3}. Multiply \frac{2}{k-3} times \frac{2\left(k+2\right)}{2\left(k+2\right)}.

$$\frac{\mathrm{d}}{\mathrm{d}k}(\frac{6\left(k-3\right)}{2\left(k-3\right)\left(k+2\right)}+\frac{2\times 2\left(k+2\right)}{2\left(k-3\right)\left(k+2\right)})$$

Since \frac{6\left(k-3\right)}{2\left(k-3\right)\left(k+2\right)} and \frac{2\times 2\left(k+2\right)}{2\left(k-3\right)\left(k+2\right)} have the same denominator, add them by adding their numerators.

$$\frac{\mathrm{d}}{\mathrm{d}k}(\frac{6\left(k-3\right)+2\times 2\left(k+2\right)}{2\left(k-3\right)\left(k+2\right)})$$

Do the multiplications in 6\left(k-3\right)+2\times 2\left(k+2\right).

$$\frac{\mathrm{d}}{\mathrm{d}k}(\frac{6k-18+4k+8}{2\left(k-3\right)\left(k+2\right)})$$

Combine like terms in 6k-18+4k+8.

$$\frac{\mathrm{d}}{\mathrm{d}k}(\frac{10k-10}{2\left(k-3\right)\left(k+2\right)})$$

Factor the expressions that are not already factored in \frac{10k-10}{2\left(k-3\right)\left(k+2\right)}.

$$\frac{\mathrm{d}}{\mathrm{d}k}(\frac{10\left(k-1\right)}{2\left(k-3\right)\left(k+2\right)})$$

Cancel out 2 in both numerator and denominator.

$$\frac{\mathrm{d}}{\mathrm{d}k}(\frac{5\left(k-1\right)}{\left(k-3\right)\left(k+2\right)})$$

Use the distributive property to multiply 5 by k-1.

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

Apply the distributive property by multiplying each term of k-3 by each term of k+2.

$$\frac{\mathrm{d}}{\mathrm{d}k}(\frac{5k-5}{k^{2}+2k-3k-6})$$

Combine 2k and -3k to get -k.

$$\frac{\mathrm{d}}{\mathrm{d}k}(\frac{5k-5}{k^{2}-k-6})$$

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.

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

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}.

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

Simplify.

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

Multiply k^{2}-k^{1}-6 times 5k^{0}.

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

Multiply 5k^{1}-5 times 2k^{1}-k^{0}.

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

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

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

Simplify.

$$\frac{5k^{2}-5k^{1}-30k^{0}-\left(10k^{2}-5k^{1}-10k^{1}+5k^{0}\right)}{\left(k^{2}-k^{1}-6\right)^{2}}$$

Combine like terms.

$$\frac{-5k^{2}+10k^{1}-35k^{0}}{\left(k^{2}-k^{1}-6\right)^{2}}$$

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

$$\frac{-5k^{2}+10k-35k^{0}}{\left(k^{2}-k-6\right)^{2}}$$

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

$$\frac{-5k^{2}+10k-35}{\left(k^{2}-k-6\right)^{2}}$$