Solve 4k/4k-1/frac{4k{2k}}

$$\frac { \frac { 4 k } { 4 k - 1 } } { \frac { 4 k } { 2 k } }$$

Evaluate

$$\frac{2k}{4k-1}$$

Solution Steps

Divide \frac{4k}{4k-1} by \frac{4k}{2k} by multiplying \frac{4k}{4k-1} by the reciprocal of \frac{4k}{2k}.

$$\frac{4k\times 2k}{\left(4k-1\right)\times 4k}$$

Cancel out 4k in both numerator and denominator.

$$\frac{2k}{4k-1}$$

Differentiate w.r.t. k

$$-\frac{2}{\left(4k-1\right)^{2}}$$

Steps Using Derivative Rule for Quotient

Divide \frac{4k}{4k-1} by \frac{4k}{2k} by multiplying \frac{4k}{4k-1} by the reciprocal of \frac{4k}{2k}.

$$\frac{\mathrm{d}}{\mathrm{d}k}(\frac{4k\times 2k}{\left(4k-1\right)\times 4k})$$

Cancel out 4k in both numerator and denominator.

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

For any two differentiable functions, the derivative of the quotient of two functions is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the denominator squared.

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

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}.

$$\frac{\left(4k^{1}-1\right)\times 2k^{1-1}-2k^{1}\times 4k^{1-1}}{\left(4k^{1}-1\right)^{2}}$$

Do the arithmetic.

$$\frac{\left(4k^{1}-1\right)\times 2k^{0}-2k^{1}\times 4k^{0}}{\left(4k^{1}-1\right)^{2}}$$

Expand using distributive property.

$$\frac{4k^{1}\times 2k^{0}-2k^{0}-2k^{1}\times 4k^{0}}{\left(4k^{1}-1\right)^{2}}$$

To multiply powers of the same base, add their exponents.

$$\frac{4\times 2k^{1}-2k^{0}-2\times 4k^{1}}{\left(4k^{1}-1\right)^{2}}$$

Do the arithmetic.

$$\frac{8k^{1}-2k^{0}-8k^{1}}{\left(4k^{1}-1\right)^{2}}$$

Combine like terms.

$$\frac{\left(8-8\right)k^{1}-2k^{0}}{\left(4k^{1}-1\right)^{2}}$$

Subtract 8 from 8.

$$\frac{-2k^{0}}{\left(4k^{1}-1\right)^{2}}$$

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

$$\frac{-2k^{0}}{\left(4k-1\right)^{2}}$$

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

$$\frac{-2}{\left(4k-1\right)^{2}}$$