Solve k/-k^3+k^2+k

$$\frac { k } { - k ^ { 3 } + k ^ { 2 } + k }$$

Evaluate

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

Short Solution Steps

Factor the expressions that are not already factored.

$$\frac{k}{-k\left(k-\left(-\frac{1}{2}\sqrt{5}+\frac{1}{2}\right)\right)\left(k-\left(\frac{1}{2}\sqrt{5}+\frac{1}{2}\right)\right)}$$

Cancel out k in both numerator and denominator.

$$\frac{1}{-\left(k-\left(-\frac{1}{2}\sqrt{5}+\frac{1}{2}\right)\right)\left(k-\left(\frac{1}{2}\sqrt{5}+\frac{1}{2}\right)\right)}$$

Expand the expression.

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

Differentiate w.r.t. k

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

Steps Using Chain Rule

Factor the expressions that are not already factored in \frac{k}{-k^{3}+k^{2}+k}.

$$\frac{\mathrm{d}}{\mathrm{d}k}(\frac{k}{-k\left(k-\left(-\frac{1}{2}\sqrt{5}+\frac{1}{2}\right)\right)\left(k-\left(\frac{1}{2}\sqrt{5}+\frac{1}{2}\right)\right)})$$

Cancel out k in both numerator and denominator.

$$\frac{\mathrm{d}}{\mathrm{d}k}(\frac{1}{-\left(k-\left(-\frac{1}{2}\sqrt{5}+\frac{1}{2}\right)\right)\left(k-\left(\frac{1}{2}\sqrt{5}+\frac{1}{2}\right)\right)})$$

To find the opposite of -\frac{1}{2}\sqrt{5}+\frac{1}{2}, find the opposite of each term.

$$\frac{\mathrm{d}}{\mathrm{d}k}(\frac{1}{-\left(k+\frac{1}{2}\sqrt{5}-\frac{1}{2}\right)\left(k-\left(\frac{1}{2}\sqrt{5}+\frac{1}{2}\right)\right)})$$

To find the opposite of \frac{1}{2}\sqrt{5}+\frac{1}{2}, find the opposite of each term.

$$\frac{\mathrm{d}}{\mathrm{d}k}(\frac{1}{-\left(k+\frac{1}{2}\sqrt{5}-\frac{1}{2}\right)\left(k-\frac{1}{2}\sqrt{5}-\frac{1}{2}\right)})$$

Use the distributive property to multiply -1 by k+\frac{1}{2}\sqrt{5}-\frac{1}{2}.

$$\frac{\mathrm{d}}{\mathrm{d}k}(\frac{1}{\left(-k-\frac{1}{2}\sqrt{5}+\frac{1}{2}\right)\left(k-\frac{1}{2}\sqrt{5}-\frac{1}{2}\right)})$$

Use the distributive property to multiply -k-\frac{1}{2}\sqrt{5}+\frac{1}{2} by k-\frac{1}{2}\sqrt{5}-\frac{1}{2} and combine like terms.

$$\frac{\mathrm{d}}{\mathrm{d}k}(\frac{1}{-k^{2}+k+\frac{1}{4}\left(\sqrt{5}\right)^{2}-\frac{1}{4}})$$

The square of \sqrt{5} is 5.

$$\frac{\mathrm{d}}{\mathrm{d}k}(\frac{1}{-k^{2}+k+\frac{1}{4}\times 5-\frac{1}{4}})$$

Multiply \frac{1}{4} and 5 to get \frac{5}{4}.

$$\frac{\mathrm{d}}{\mathrm{d}k}(\frac{1}{-k^{2}+k+\frac{5}{4}-\frac{1}{4}})$$

Subtract \frac{1}{4} from \frac{5}{4} to get 1.

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

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).

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

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}.

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

Simplify.

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

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

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

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

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