Solve 3x/x^2-25+2/2x+10

$$\frac { 3 x } { x ^ { 2 } - 25 } + \frac { 2 } { 2 x + 10 }$$

Evaluate

$$\frac{4x-5}{x^{2}-25}$$

Short Solution Steps

Factor x^{2}-25. Factor 2x+10.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x-5\right)\left(x+5\right) and 2\left(x+5\right) is 2\left(x-5\right)\left(x+5\right). Multiply \frac{3x}{\left(x-5\right)\left(x+5\right)} times \frac{2}{2}. Multiply \frac{2}{2\left(x+5\right)} times \frac{x-5}{x-5}.

$$\frac{2\times 3x}{2\left(x-5\right)\left(x+5\right)}+\frac{2\left(x-5\right)}{2\left(x-5\right)\left(x+5\right)}$$

Since \frac{2\times 3x}{2\left(x-5\right)\left(x+5\right)} and \frac{2\left(x-5\right)}{2\left(x-5\right)\left(x+5\right)} have the same denominator, add them by adding their numerators.

$$\frac{2\times 3x+2\left(x-5\right)}{2\left(x-5\right)\left(x+5\right)}$$

Do the multiplications in 2\times 3x+2\left(x-5\right).

$$\frac{6x+2x-10}{2\left(x-5\right)\left(x+5\right)}$$

Combine like terms in 6x+2x-10.

$$\frac{8x-10}{2\left(x-5\right)\left(x+5\right)}$$

Factor the expressions that are not already factored in \frac{8x-10}{2\left(x-5\right)\left(x+5\right)}.

$$\frac{2\left(4x-5\right)}{2\left(x-5\right)\left(x+5\right)}$$

Cancel out 2 in both numerator and denominator.

$$\frac{4x-5}{\left(x-5\right)\left(x+5\right)}$$

Expand \left(x-5\right)\left(x+5\right).

$$\frac{4x-5}{x^{2}-25}$$

Differentiate w.r.t. x

$$\frac{2\left(-2x^{2}+5x-50\right)}{\left(x^{2}-25\right)^{2}}$$

Steps Using Derivative Rule for Quotient

Factor x^{2}-25. Factor 2x+10.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x-5\right)\left(x+5\right) and 2\left(x+5\right) is 2\left(x-5\right)\left(x+5\right). Multiply \frac{3x}{\left(x-5\right)\left(x+5\right)} times \frac{2}{2}. Multiply \frac{2}{2\left(x+5\right)} times \frac{x-5}{x-5}.

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

Since \frac{2\times 3x}{2\left(x-5\right)\left(x+5\right)} and \frac{2\left(x-5\right)}{2\left(x-5\right)\left(x+5\right)} have the same denominator, add them by adding their numerators.

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

Do the multiplications in 2\times 3x+2\left(x-5\right).

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{6x+2x-10}{2\left(x-5\right)\left(x+5\right)})$$

Combine like terms in 6x+2x-10.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{8x-10}{2\left(x-5\right)\left(x+5\right)})$$

Factor the expressions that are not already factored in \frac{8x-10}{2\left(x-5\right)\left(x+5\right)}.

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

Cancel out 2 in both numerator and denominator.

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

Consider \left(x-5\right)\left(x+5\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 5.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{4x-5}{x^{2}-25})$$

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.

$$\frac{\left(x^{2}-25\right)\frac{\mathrm{d}}{\mathrm{d}x}(4x^{1}-5)-\left(4x^{1}-5\right)\frac{\mathrm{d}}{\mathrm{d}x}(x^{2}-25)}{\left(x^{2}-25\right)^{2}}$$

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}.

$$\frac{\left(x^{2}-25\right)\times 4x^{1-1}-\left(4x^{1}-5\right)\times 2x^{2-1}}{\left(x^{2}-25\right)^{2}}$$

Do the arithmetic.

$$\frac{\left(x^{2}-25\right)\times 4x^{0}-\left(4x^{1}-5\right)\times 2x^{1}}{\left(x^{2}-25\right)^{2}}$$

Expand using distributive property.

$$\frac{x^{2}\times 4x^{0}-25\times 4x^{0}-\left(4x^{1}\times 2x^{1}-5\times 2x^{1}\right)}{\left(x^{2}-25\right)^{2}}$$

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

$$\frac{4x^{2}-25\times 4x^{0}-\left(4\times 2x^{1+1}-5\times 2x^{1}\right)}{\left(x^{2}-25\right)^{2}}$$

Do the arithmetic.

$$\frac{4x^{2}-100x^{0}-\left(8x^{2}-10x^{1}\right)}{\left(x^{2}-25\right)^{2}}$$

Remove unnecessary parentheses.

$$\frac{4x^{2}-100x^{0}-8x^{2}-\left(-10x^{1}\right)}{\left(x^{2}-25\right)^{2}}$$

Combine like terms.

$$\frac{\left(4-8\right)x^{2}-100x^{0}-\left(-10x^{1}\right)}{\left(x^{2}-25\right)^{2}}$$

Subtract 8 from 4.

$$\frac{-4x^{2}-100x^{0}-\left(-10x^{1}\right)}{\left(x^{2}-25\right)^{2}}$$

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

$$\frac{-4x^{2}-100x^{0}-\left(-10x\right)}{\left(x^{2}-25\right)^{2}}$$

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

$$\frac{-4x^{2}-100-\left(-10x\right)}{\left(x^{2}-25\right)^{2}}$$