Calculus Examples

$(\frac{b^{2} + z^{2}}{b^{2} - z^{2}}) (- \frac{4 b^{2} z}{{(b^{2} + z^{2})}^{2}})$

Step 1

Simplify terms.

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Step 1.1

Rewrite using the commutative property of multiplication.

$- \frac{b^{2} + z^{2}}{b^{2} - z^{2}} \cdot \frac{4 b^{2} z}{{(b^{2} + z^{2})}^{2}}$

Step 1.2

Cancel the common factor of $b^{2} + z^{2}$ .

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Step 1.2.1

Move the leading negative in $- \frac{b^{2} + z^{2}}{b^{2} - z^{2}}$ into the numerator.

$\frac{- (b^{2} + z^{2})}{b^{2} - z^{2}} \cdot \frac{4 b^{2} z}{{(b^{2} + z^{2})}^{2}}$

Step 1.2.2

Factor $b^{2} + z^{2}$ out of $- (b^{2} + z^{2})$ .

$\frac{(b^{2} + z^{2}) \cdot - 1}{b^{2} - z^{2}} \cdot \frac{4 b^{2} z}{{(b^{2} + z^{2})}^{2}}$

Step 1.2.3

Factor $b^{2} + z^{2}$ out of ${(b^{2} + z^{2})}^{2}$ .

$\frac{(b^{2} + z^{2}) \cdot - 1}{b^{2} - z^{2}} \cdot \frac{4 b^{2} z}{(b^{2} + z^{2}) (b^{2} + z^{2})}$

Step 1.2.4

Cancel the common factor.

$\frac{(b^{2} + z^{2}) \cdot - 1}{b^{2} - z^{2}} \cdot \frac{4 b^{2} z}{(b^{2} + z^{2}) (b^{2} + z^{2})}$

Step 1.2.5

Rewrite the expression.

$\frac{- 1}{b^{2} - z^{2}} \cdot \frac{4 b^{2} z}{b^{2} + z^{2}}$

Step 1.3

Multiply $\frac{- 1}{b^{2} - z^{2}}$ by $\frac{4 b^{2} z}{b^{2} + z^{2}}$ .

$\frac{- (4 b^{2} z)}{(b^{2} - z^{2}) (b^{2} + z^{2})}$

Step 1.4

Multiply $4$ by $- 1$ .

$\frac{- 4 (b^{2} z)}{(b^{2} - z^{2}) (b^{2} + z^{2})}$

Step 2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = b$ and $b = z$ .

$\frac{- 4 b^{2} z}{(b + z) (b - z) (b^{2} + z^{2})}$

Step 3

Move the negative in front of the fraction.

$- \frac{4 b^{2} z}{(b + z) (b - z) (b^{2} + z^{2})}$