Solve 1/k+1(k+1)=

$$\frac { 1 } { k + 1 ( k + 1 ) } =$$

Evaluate

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

Solution Steps

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

$$\frac{1}{k+k+1}$$

Combine k and k to get 2k.

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

Differentiate w.r.t. k

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

Steps Using Chain Rule

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

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

Combine k and k to get 2k.

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

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).

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

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}.

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

Simplify.

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

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

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

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

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