Solve 5/frac{7{x}+2}

$$\frac { 5 } { \frac { 7 } { x } + 2 }$$

Evaluate

$$\frac{5x}{2x+7}$$

Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{x}{x}.

$$\frac{5}{\frac{7}{x}+\frac{2x}{x}}$$

Since \frac{7}{x} and \frac{2x}{x} have the same denominator, add them by adding their numerators.

$$\frac{5}{\frac{7+2x}{x}}$$

Divide 5 by \frac{7+2x}{x} by multiplying 5 by the reciprocal of \frac{7+2x}{x}.

$$\frac{5x}{7+2x}$$

Differentiate w.r.t. x

$$\frac{35}{\left(2x+7\right)^{2}}$$

Steps Using Derivative Rule for Quotient

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{x}{x}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{5}{\frac{7}{x}+\frac{2x}{x}})$$

Since \frac{7}{x} and \frac{2x}{x} have the same denominator, add them by adding their numerators.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{5}{\frac{7+2x}{x}})$$

Divide 5 by \frac{7+2x}{x} by multiplying 5 by the reciprocal of \frac{7+2x}{x}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{5x}{7+2x})$$

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.

$$\frac{\left(2x^{1}+7\right)\frac{\mathrm{d}}{\mathrm{d}x}(5x^{1})-5x^{1}\frac{\mathrm{d}}{\mathrm{d}x}(2x^{1}+7)}{\left(2x^{1}+7\right)^{2}}$$

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}.

$$\frac{\left(2x^{1}+7\right)\times 5x^{1-1}-5x^{1}\times 2x^{1-1}}{\left(2x^{1}+7\right)^{2}}$$

Do the arithmetic.

$$\frac{\left(2x^{1}+7\right)\times 5x^{0}-5x^{1}\times 2x^{0}}{\left(2x^{1}+7\right)^{2}}$$

Expand using distributive property.

$$\frac{2x^{1}\times 5x^{0}+7\times 5x^{0}-5x^{1}\times 2x^{0}}{\left(2x^{1}+7\right)^{2}}$$

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

$$\frac{2\times 5x^{1}+7\times 5x^{0}-5\times 2x^{1}}{\left(2x^{1}+7\right)^{2}}$$

Do the arithmetic.

$$\frac{10x^{1}+35x^{0}-10x^{1}}{\left(2x^{1}+7\right)^{2}}$$

Combine like terms.

$$\frac{\left(10-10\right)x^{1}+35x^{0}}{\left(2x^{1}+7\right)^{2}}$$

Subtract 10 from 10.

$$\frac{35x^{0}}{\left(2x^{1}+7\right)^{2}}$$

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

$$\frac{35x^{0}}{\left(2x+7\right)^{2}}$$

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

$$\frac{35\times 1}{\left(2x+7\right)^{2}}$$

For any term t, t\times 1=t and 1t=t.

$$\frac{35}{\left(2x+7\right)^{2}}$$