Solve a+b-a/1+ab/1-frac{ab-a^2{1+ab}}

$$\frac { a + \frac { b - a } { 1 + a b } } { 1 - \frac { a b - a ^ { 2 } } { 1 + a b } }$$

Evaluate

$$b$$

Short Solution Steps

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

$$\frac{\frac{a\left(1+ab\right)}{1+ab}+\frac{b-a}{1+ab}}{1-\frac{ab-a^{2}}{1+ab}}$$

Since \frac{a\left(1+ab\right)}{1+ab} and \frac{b-a}{1+ab} have the same denominator, add them by adding their numerators.

$$\frac{\frac{a\left(1+ab\right)+b-a}{1+ab}}{1-\frac{ab-a^{2}}{1+ab}}$$

Do the multiplications in a\left(1+ab\right)+b-a.

$$\frac{\frac{a+a^{2}b+b-a}{1+ab}}{1-\frac{ab-a^{2}}{1+ab}}$$

Combine like terms in a+a^{2}b+b-a.

$$\frac{\frac{b+a^{2}b}{1+ab}}{1-\frac{ab-a^{2}}{1+ab}}$$

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

$$\frac{\frac{b+a^{2}b}{1+ab}}{\frac{1+ab}{1+ab}-\frac{ab-a^{2}}{1+ab}}$$

Since \frac{1+ab}{1+ab} and \frac{ab-a^{2}}{1+ab} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{b+a^{2}b}{1+ab}}{\frac{1+ab-\left(ab-a^{2}\right)}{1+ab}}$$

Do the multiplications in 1+ab-\left(ab-a^{2}\right).

$$\frac{\frac{b+a^{2}b}{1+ab}}{\frac{1+ab-ab+a^{2}}{1+ab}}$$

Combine like terms in 1+ab-ab+a^{2}.

$$\frac{\frac{b+a^{2}b}{1+ab}}{\frac{1+a^{2}}{1+ab}}$$

Divide \frac{b+a^{2}b}{1+ab} by \frac{1+a^{2}}{1+ab} by multiplying \frac{b+a^{2}b}{1+ab} by the reciprocal of \frac{1+a^{2}}{1+ab}.

$$\frac{\left(b+a^{2}b\right)\left(1+ab\right)}{\left(1+ab\right)\left(1+a^{2}\right)}$$

Cancel out ab+1 in both numerator and denominator.

$$\frac{ba^{2}+b}{a^{2}+1}$$

Factor the expressions that are not already factored.

$$\frac{b\left(a^{2}+1\right)}{a^{2}+1}$$

Cancel out a^{2}+1 in both numerator and denominator.

$$b$$

Differentiate w.r.t. b

$$1$$

Steps Using Definition of a Derivative

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

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

Since \frac{a\left(1+ab\right)}{1+ab} and \frac{b-a}{1+ab} have the same denominator, add them by adding their numerators.

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

Do the multiplications in a\left(1+ab\right)+b-a.

$$\frac{\mathrm{d}}{\mathrm{d}b}(\frac{\frac{a+a^{2}b+b-a}{1+ab}}{1-\frac{ab-a^{2}}{1+ab}})$$

Combine like terms in a+a^{2}b+b-a.

$$\frac{\mathrm{d}}{\mathrm{d}b}(\frac{\frac{b+a^{2}b}{1+ab}}{1-\frac{ab-a^{2}}{1+ab}})$$

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

$$\frac{\mathrm{d}}{\mathrm{d}b}(\frac{\frac{b+a^{2}b}{1+ab}}{\frac{1+ab}{1+ab}-\frac{ab-a^{2}}{1+ab}})$$

Since \frac{1+ab}{1+ab} and \frac{ab-a^{2}}{1+ab} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 1+ab-\left(ab-a^{2}\right).

$$\frac{\mathrm{d}}{\mathrm{d}b}(\frac{\frac{b+a^{2}b}{1+ab}}{\frac{1+ab-ab+a^{2}}{1+ab}})$$

Combine like terms in 1+ab-ab+a^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}b}(\frac{\frac{b+a^{2}b}{1+ab}}{\frac{1+a^{2}}{1+ab}})$$

Divide \frac{b+a^{2}b}{1+ab} by \frac{1+a^{2}}{1+ab} by multiplying \frac{b+a^{2}b}{1+ab} by the reciprocal of \frac{1+a^{2}}{1+ab}.

$$\frac{\mathrm{d}}{\mathrm{d}b}(\frac{\left(b+a^{2}b\right)\left(1+ab\right)}{\left(1+ab\right)\left(1+a^{2}\right)})$$

Cancel out ab+1 in both numerator and denominator.

$$\frac{\mathrm{d}}{\mathrm{d}b}(\frac{ba^{2}+b}{a^{2}+1})$$

Factor the expressions that are not already factored in \frac{ba^{2}+b}{a^{2}+1}.

$$\frac{\mathrm{d}}{\mathrm{d}b}(\frac{b\left(a^{2}+1\right)}{a^{2}+1})$$

Cancel out a^{2}+1 in both numerator and denominator.

$$\frac{\mathrm{d}}{\mathrm{d}b}(b)$$

The derivative of ax^{n} is nax^{n-1}.

$$b^{1-1}$$

Subtract 1 from 1.

$$b^{0}$$

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

$$1$$