Solve 1/3x+1-1/2x+1

$$\frac { 1 } { 3 x + 1 } - \frac { 1 } { 2 x + 1 }$$

Evaluate

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

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3x+1 and 2x+1 is \left(2x+1\right)\left(3x+1\right). Multiply \frac{1}{3x+1} times \frac{2x+1}{2x+1}. Multiply \frac{1}{2x+1} times \frac{3x+1}{3x+1}.

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

Since \frac{2x+1}{\left(2x+1\right)\left(3x+1\right)} and \frac{3x+1}{\left(2x+1\right)\left(3x+1\right)} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 2x+1-\left(3x+1\right).

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

Combine like terms in 2x+1-3x-1.

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

Expand \left(2x+1\right)\left(3x+1\right).

$$\frac{-x}{6x^{2}+5x+1}$$

Differentiate w.r.t. x

$$\frac{6x^{2}-1}{\left(\left(2x+1\right)\left(3x+1\right)\right)^{2}}$$

Steps Using Derivative Rule for Quotient

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3x+1 and 2x+1 is \left(2x+1\right)\left(3x+1\right). Multiply \frac{1}{3x+1} times \frac{2x+1}{2x+1}. Multiply \frac{1}{2x+1} times \frac{3x+1}{3x+1}.

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

Since \frac{2x+1}{\left(2x+1\right)\left(3x+1\right)} and \frac{3x+1}{\left(2x+1\right)\left(3x+1\right)} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 2x+1-\left(3x+1\right).

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

Combine like terms in 2x+1-3x-1.

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

Apply the distributive property by multiplying each term of 2x+1 by each term of 3x+1.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{-x}{6x^{2}+2x+3x+1})$$

Combine 2x and 3x to get 5x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{-x}{6x^{2}+5x+1})$$

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.

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

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}.

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

Simplify.

$$\frac{\left(6x^{2}+5x^{1}+1\right)\left(-1\right)x^{0}-\left(-x^{1}\left(12x^{1}+5x^{0}\right)\right)}{\left(6x^{2}+5x^{1}+1\right)^{2}}$$

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

$$\frac{6x^{2}\left(-1\right)x^{0}+5x^{1}\left(-1\right)x^{0}-x^{0}-\left(-x^{1}\left(12x^{1}+5x^{0}\right)\right)}{\left(6x^{2}+5x^{1}+1\right)^{2}}$$

Multiply -x^{1} times 12x^{1}+5x^{0}.

$$\frac{6x^{2}\left(-1\right)x^{0}+5x^{1}\left(-1\right)x^{0}-x^{0}-\left(-x^{1}\times 12x^{1}-x^{1}\times 5x^{0}\right)}{\left(6x^{2}+5x^{1}+1\right)^{2}}$$

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

$$\frac{6\left(-1\right)x^{2}+5\left(-1\right)x^{1}-x^{0}-\left(-12x^{1+1}-5x^{1}\right)}{\left(6x^{2}+5x^{1}+1\right)^{2}}$$

Simplify.

$$\frac{-6x^{2}-5x^{1}-x^{0}-\left(-12x^{2}-5x^{1}\right)}{\left(6x^{2}+5x^{1}+1\right)^{2}}$$

Combine like terms.

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

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

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

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

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