Solve 1/1+x/2

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

Evaluate

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

Solution Steps

Express \frac{\frac{1}{1+x}}{2} as a single fraction.

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

Use the distributive property to multiply 1+x by 2.

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

Differentiate w.r.t. x

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

Steps Using Chain Rule

Express \frac{\frac{1}{1+x}}{2} as a single fraction.

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

Use the distributive property to multiply 1+x by 2.

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

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).

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

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}.

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

Simplify.

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

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

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

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

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