Calculus Examples

$y = 19 z^{2} + \frac{1}{5 z}$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d y} (y) = \frac{d}{d y} (19 z^{2} + \frac{1}{5 z})$

Step 2

Differentiate using the Power Rule which states that $\frac{d}{d y} [y^{n}]$ is $n y^{n - 1}$ where $n = 1$ .

$1$

Step 3

Differentiate the right side of the equation.

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Step 3.1

By the Sum Rule, the derivative of $19 z^{2} + \frac{1}{5 z}$ with respect to $y$ is $\frac{d}{d y} [19 z^{2}] + \frac{d}{d y} [\frac{1}{5 z}]$ .

$\frac{d}{d y} [19 z^{2}] + \frac{d}{d y} [\frac{1}{5 z}]$

Step 3.2

Evaluate $\frac{d}{d y} [19 z^{2}]$ .

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Step 3.2.1

Since $19$ is constant with respect to $y$ , the derivative of $19 z^{2}$ with respect to $y$ is $19 \frac{d}{d y} [z^{2}]$ .

$19 \frac{d}{d y} [z^{2}] + \frac{d}{d y} [\frac{1}{5 z}]$

Step 3.2.2

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = y^{2}$ and $g (y) = z$ .

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Step 3.2.2.1

To apply the Chain Rule, set $u$ as $z$ .

$19 (\frac{d}{d u} [u^{2}] \frac{d}{d y} [z]) + \frac{d}{d y} [\frac{1}{5 z}]$

Step 3.2.2.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 2$ .

$19 (2 u \frac{d}{d y} [z]) + \frac{d}{d y} [\frac{1}{5 z}]$

Step 3.2.2.3

Replace all occurrences of $u$ with $z$ .

$19 (2 z \frac{d}{d y} [z]) + \frac{d}{d y} [\frac{1}{5 z}]$

Step 3.2.3

Rewrite $\frac{d}{d y} [z]$ as $z'$ .

$19 (2 z z') + \frac{d}{d y} [\frac{1}{5 z}]$

Step 3.2.4

Multiply $2$ by $19$ .

$38 z z' + \frac{d}{d y} [\frac{1}{5 z}]$

Step 3.3

Evaluate $\frac{d}{d y} [\frac{1}{5 z}]$ .

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Step 3.3.1

Since $\frac{1}{5}$ is constant with respect to $y$ , the derivative of $\frac{1}{5 z}$ with respect to $y$ is $\frac{1}{5} \frac{d}{d y} [\frac{1}{z}]$ .

$38 z z' + \frac{1}{5} \frac{d}{d y} [\frac{1}{z}]$

Step 3.3.2

Differentiate using the Quotient Rule which states that $\frac{d}{d y} [\frac{f (y)}{g (y)}]$ is $\frac{g (y) \frac{d}{d y} [f (y)] - f (y) \frac{d}{d y} [g (y)]}{{g (y)}^{2}}$ where $f (y) = 1$ and $g (y) = z$ .

$38 z z' + \frac{1}{5} \cdot \frac{z \frac{d}{d y} [1] - 1 \cdot 1 \frac{d}{d y} [z]}{z^{2}}$

Step 3.3.3

Since $1$ is constant with respect to $y$ , the derivative of $1$ with respect to $y$ is $0$ .

$38 z z' + \frac{1}{5} \cdot \frac{z \cdot 0 - 1 \cdot 1 \frac{d}{d y} [z]}{z^{2}}$

Step 3.3.4

Rewrite $\frac{d}{d y} [z]$ as $z'$ .

$38 z z' + \frac{1}{5} \cdot \frac{z \cdot 0 - 1 \cdot 1 z'}{z^{2}}$

Step 3.3.5

Multiply $z$ by $0$ .

$38 z z' + \frac{1}{5} \cdot \frac{0 - 1 \cdot 1 z'}{z^{2}}$

Step 3.3.6

Multiply $- 1$ by $1$ .

$38 z z' + \frac{1}{5} \cdot \frac{0 - z'}{z^{2}}$

Step 3.3.7

Subtract $z'$ from $0$ .

$38 z z' + \frac{1}{5} \cdot \frac{- z'}{z^{2}}$

Step 3.3.8

Move the negative in front of the fraction.

$38 z z' + \frac{1}{5} (- \frac{z'}{z^{2}})$

Step 3.3.9

Multiply $\frac{1}{5}$ by $\frac{z'}{z^{2}}$ .

$38 z z' - \frac{z'}{5 z^{2}}$

Step 4

Reform the equation by setting the left side equal to the right side.

$1 = 38 z z' - \frac{z'}{5 z^{2}}$

Step 5

Solve for $z'$ .

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Step 5.1

Rewrite the equation as $38 z z' - \frac{z'}{5 z^{2}} = 1$ .

$38 z z' - \frac{z'}{5 z^{2}} = 1$

Step 5.2

Find the LCD of the terms in the equation.

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Step 5.2.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$1, 5 z^{2}, 1$

Step 5.2.2

The LCM of one and any expression is the expression.

$5 z^{2}$

Step 5.3

Multiply each term in $38 z z' - \frac{z'}{5 z^{2}} = 1$ by $5 z^{2}$ to eliminate the fractions.

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Step 5.3.1

Multiply each term in $38 z z' - \frac{z'}{5 z^{2}} = 1$ by $5 z^{2}$ .

$38 z z' (5 z^{2}) - \frac{z'}{5 z^{2}} (5 z^{2}) = 1 (5 z^{2})$

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Simplify each term.

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Step 5.3.2.1.1

Multiply $z$ by $z^{2}$ by adding the exponents.

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Step 5.3.2.1.1.1

Move $z^{2}$ .

$38 (z^{2} z) z' \cdot 5 - \frac{z'}{5 z^{2}} (5 z^{2}) = 1 (5 z^{2})$

Step 5.3.2.1.1.2

Multiply $z^{2}$ by $z$ .

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Step 5.3.2.1.1.2.1

Raise $z$ to the power of $1$ .

$38 (z^{2} z^{1}) z' \cdot 5 - \frac{z'}{5 z^{2}} (5 z^{2}) = 1 (5 z^{2})$

Step 5.3.2.1.1.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$38 z^{2 + 1} z' \cdot 5 - \frac{z'}{5 z^{2}} (5 z^{2}) = 1 (5 z^{2})$

Step 5.3.2.1.1.3

Add $2$ and $1$ .

$38 z^{3} z' \cdot 5 - \frac{z'}{5 z^{2}} (5 z^{2}) = 1 (5 z^{2})$

Step 5.3.2.1.2

Multiply $5$ by $38$ .

$190 z^{3} z' - \frac{z'}{5 z^{2}} (5 z^{2}) = 1 (5 z^{2})$

Step 5.3.2.1.3

Cancel the common factor of $5 z^{2}$ .

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Step 5.3.2.1.3.1

Move the leading negative in $- \frac{z'}{5 z^{2}}$ into the numerator.

$190 z^{3} z' + \frac{- z'}{5 z^{2}} (5 z^{2}) = 1 (5 z^{2})$

Step 5.3.2.1.3.2

Cancel the common factor.

$190 z^{3} z' + \frac{- z'}{5 z^{2}} (5 z^{2}) = 1 (5 z^{2})$

Step 5.3.2.1.3.3

Rewrite the expression.

$190 z^{3} z' - z' = 1 (5 z^{2})$

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Multiply $5 z^{2}$ by $1$ .

$190 z^{3} z' - z' = 5 z^{2}$

Step 5.4

Solve the equation.

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Step 5.4.1

Factor $z'$ out of $190 z^{3} z' - z'$ .

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Step 5.4.1.1

Factor $z'$ out of $190 z^{3} z'$ .

$z' (190 z^{3}) - z' = 5 z^{2}$

Step 5.4.1.2

Factor $z'$ out of $- z'$ .

$z' (190 z^{3}) + z' \cdot - 1 = 5 z^{2}$

Step 5.4.1.3

Factor $z'$ out of $z' (190 z^{3}) + z' \cdot - 1$ .

$z' (190 z^{3} - 1) = 5 z^{2}$

Step 5.4.2

Divide each term in $z' (190 z^{3} - 1) = 5 z^{2}$ by $190 z^{3} - 1$ and simplify.

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Step 5.4.2.1

Divide each term in $z' (190 z^{3} - 1) = 5 z^{2}$ by $190 z^{3} - 1$ .

$\frac{z' (190 z^{3} - 1)}{190 z^{3} - 1} = \frac{5 z^{2}}{190 z^{3} - 1}$

Step 5.4.2.2

Simplify the left side.

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Step 5.4.2.2.1

Cancel the common factor of $190 z^{3} - 1$ .

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Step 5.4.2.2.1.1

Cancel the common factor.

$\frac{z' (190 z^{3} - 1)}{190 z^{3} - 1} = \frac{5 z^{2}}{190 z^{3} - 1}$

Step 5.4.2.2.1.2

Divide $z'$ by $1$ .

$z' = \frac{5 z^{2}}{190 z^{3} - 1}$

Step 6

Replace $z'$ with $\frac{d z}{d y}$ .

$\frac{d z}{d y} = \frac{5 z^{2}}{190 z^{3} - 1}$