Solve -1/x+1*x+1

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

Evaluate

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

Short Solution Steps

Express \left(-\frac{1}{x+1}\right)x as a single fraction.

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

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

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

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

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

Combine like terms in -x+x+1.

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

Differentiate w.r.t. x

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

Steps Using Chain Rule

Express \left(-\frac{1}{x+1}\right)x as a single fraction.

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

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

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

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

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

Combine like terms in -x+x+1.

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

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.

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