Solve (2x+3/2x-3-2x-3/2x+3):24/4x^2-9

$$( \frac { 2 x + 3 } { 2 x - 3 } - \frac { 2 x - 3 } { 2 x + 3 } ) : \frac { 24 } { 4 x ^ { 2 } - 9 }$$

Evaluate

$$x$$

Short Solution Steps

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

$$\frac{\frac{\left(2x+3\right)\left(2x+3\right)}{\left(2x-3\right)\left(2x+3\right)}-\frac{\left(2x-3\right)\left(2x-3\right)}{\left(2x-3\right)\left(2x+3\right)}}{\frac{24}{4x^{2}-9}}$$

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

$$\frac{\frac{\left(2x+3\right)\left(2x+3\right)-\left(2x-3\right)\left(2x-3\right)}{\left(2x-3\right)\left(2x+3\right)}}{\frac{24}{4x^{2}-9}}$$

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

$$\frac{\frac{4x^{2}+6x+6x+9-4x^{2}+6x+6x-9}{\left(2x-3\right)\left(2x+3\right)}}{\frac{24}{4x^{2}-9}}$$

Combine like terms in 4x^{2}+6x+6x+9-4x^{2}+6x+6x-9.

$$\frac{\frac{24x}{\left(2x-3\right)\left(2x+3\right)}}{\frac{24}{4x^{2}-9}}$$

Divide \frac{24x}{\left(2x-3\right)\left(2x+3\right)} by \frac{24}{4x^{2}-9} by multiplying \frac{24x}{\left(2x-3\right)\left(2x+3\right)} by the reciprocal of \frac{24}{4x^{2}-9}.

$$\frac{24x\left(4x^{2}-9\right)}{\left(2x-3\right)\left(2x+3\right)\times 24}$$

Cancel out 24 in both numerator and denominator.

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

Factor the expressions that are not already factored.

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

Cancel out \left(2x-3\right)\left(2x+3\right) in both numerator and denominator.

$$x$$

Differentiate w.r.t. x

$$1$$

Steps Using Definition of a Derivative

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

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{\frac{\left(2x+3\right)\left(2x+3\right)}{\left(2x-3\right)\left(2x+3\right)}-\frac{\left(2x-3\right)\left(2x-3\right)}{\left(2x-3\right)\left(2x+3\right)}}{\frac{24}{4x^{2}-9}})$$

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

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{\frac{\left(2x+3\right)\left(2x+3\right)-\left(2x-3\right)\left(2x-3\right)}{\left(2x-3\right)\left(2x+3\right)}}{\frac{24}{4x^{2}-9}})$$

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

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{\frac{4x^{2}+6x+6x+9-4x^{2}+6x+6x-9}{\left(2x-3\right)\left(2x+3\right)}}{\frac{24}{4x^{2}-9}})$$

Combine like terms in 4x^{2}+6x+6x+9-4x^{2}+6x+6x-9.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{\frac{24x}{\left(2x-3\right)\left(2x+3\right)}}{\frac{24}{4x^{2}-9}})$$

Divide \frac{24x}{\left(2x-3\right)\left(2x+3\right)} by \frac{24}{4x^{2}-9} by multiplying \frac{24x}{\left(2x-3\right)\left(2x+3\right)} by the reciprocal of \frac{24}{4x^{2}-9}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{24x\left(4x^{2}-9\right)}{\left(2x-3\right)\left(2x+3\right)\times 24})$$

Cancel out 24 in both numerator and denominator.

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

Factor the expressions that are not already factored in \frac{x\left(4x^{2}-9\right)}{\left(2x-3\right)\left(2x+3\right)}.

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

Cancel out \left(2x-3\right)\left(2x+3\right) in both numerator and denominator.

$$\frac{\mathrm{d}}{\mathrm{d}x}(x)$$

The derivative of ax^{n} is nax^{n-1}.

$$x^{1-1}$$

Subtract 1 from 1.

$$x^{0}$$

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

$$1$$