Solve 1-1/1+frac{1{x}}

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

Evaluate

$$\frac{1}{x+1}$$

Short Solution Steps

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

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

Since \frac{x}{x} and \frac{1}{x} have the same denominator, add them by adding their numerators.

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

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

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

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

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

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

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

Combine like terms in x+1-x.

$$\frac{1}{x+1}$$

Differentiate w.r.t. x

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

Steps Using Chain Rule

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

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

Since \frac{x}{x} and \frac{1}{x} have the same denominator, add them by adding their numerators.

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

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

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

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

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

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

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

Combine like terms in x+1-x.

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

If F is the composition of two differentiable functions f\left(u\right) and u=g\left(x\right), that is, if F\left(x\right)=f\left(g\left(x\right)\right), then the derivative of F is the derivative of f with respect to u times the derivative of g with respect to x, that is, \frac{\mathrm{d}}{\mathrm{d}x}(F)\left(x\right)=\frac{\mathrm{d}}{\mathrm{d}x}(f)\left(g\left(x\right)\right)\frac{\mathrm{d}}{\mathrm{d}x}(g)\left(x\right).

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

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}.

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

Simplify.

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

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

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

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

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