Solve x^2-(x/1-frac{x{x+1}})

$$x ^ { 2 } - ( \frac { x } { 1 - \frac { x } { x + 1 } } )$$

Evaluate

$$-x$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{x+1}{x+1}.

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

Since \frac{x+1}{x+1} and \frac{x}{x+1} have the same denominator, subtract them by subtracting their numerators.

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

Combine like terms in x+1-x.

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

Divide x by \frac{1}{x+1} by multiplying x by the reciprocal of \frac{1}{x+1}.

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

Use the distributive property to multiply x by x+1.

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

To find the opposite of x^{2}+x, find the opposite of each term.

$$x^{2}-x^{2}-x$$

Combine x^{2} and -x^{2} to get 0.

$$-x$$

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. Multiply 1 times \frac{x+1}{x+1}.

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

Since \frac{x+1}{x+1} and \frac{x}{x+1} have the same denominator, subtract them by subtracting their numerators.

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

Combine like terms in x+1-x.

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

Divide x by \frac{1}{x+1} by multiplying x by the reciprocal of \frac{1}{x+1}.

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

Use the distributive property to multiply x by x+1.

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

To find the opposite of x^{2}+x, find the opposite of each term.

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

Combine x^{2} and -x^{2} to get 0.

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

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.

$$-1$$