Calculus Examples

$z = \frac{A}{y^{9}} + B e^{y}$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d B} (z) = \frac{d}{d B} (\frac{A}{y^{9}} + B e^{y})$

Step 2

Since $z$ is constant with respect to $B$ , the derivative of $z$ with respect to $B$ is $0$ .

$0$

Step 3

Differentiate the right side of the equation.

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Step 3.1

By the Sum Rule, the derivative of $\frac{A}{y^{9}} + B e^{y}$ with respect to $B$ is $\frac{d}{d B} [\frac{A}{y^{9}}] + \frac{d}{d B} [B e^{y}]$ .

$\frac{d}{d B} [\frac{A}{y^{9}}] + \frac{d}{d B} [B e^{y}]$

Step 3.2

Evaluate $\frac{d}{d B} [\frac{A}{y^{9}}]$ .

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Step 3.2.1

Since $A$ is constant with respect to $B$ , the derivative of $\frac{A}{y^{9}}$ with respect to $B$ is $A \frac{d}{d B} [\frac{1}{y^{9}}]$ .

$A \frac{d}{d B} [\frac{1}{y^{9}}] + \frac{d}{d B} [B e^{y}]$

Step 3.2.2

Differentiate using the Quotient Rule which states that $\frac{d}{d B} [\frac{f (B)}{g (B)}]$ is $\frac{g (B) \frac{d}{d B} [f (B)] - f (B) \frac{d}{d B} [g (B)]}{{g (B)}^{2}}$ where $f (B) = 1$ and $g (B) = y^{9}$ .

$A \frac{y^{9} \frac{d}{d B} [1] - 1 \cdot 1 \frac{d}{d B} [y^{9}]}{{(y^{9})}^{2}} + \frac{d}{d B} [B e^{y}]$

Step 3.2.3

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

$A \frac{y^{9} \cdot 0 - 1 \cdot 1 \frac{d}{d B} [y^{9}]}{{(y^{9})}^{2}} + \frac{d}{d B} [B e^{y}]$

Step 3.2.4

Differentiate using the chain rule, which states that $\frac{d}{d B} [f (g (B))]$ is $f' (g (B)) g' (B)$ where $f (B) = B^{9}$ and $g (B) = y$ .

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Step 3.2.4.1

To apply the Chain Rule, set $u_{1}$ as $y$ .

$A \frac{y^{9} \cdot 0 - 1 \cdot 1 (\frac{d}{d u_{1}} [{u_{1}}^{9}] \frac{d}{d B} [y])}{{(y^{9})}^{2}} + \frac{d}{d B} [B e^{y}]$

Step 3.2.4.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{1}} [{u_{1}}^{n}]$ is $n {u_{1}}^{n - 1}$ where $n = 9$ .

$A \frac{y^{9} \cdot 0 - 1 \cdot 1 (9 {u_{1}}^{8} \frac{d}{d B} [y])}{{(y^{9})}^{2}} + \frac{d}{d B} [B e^{y}]$

Step 3.2.4.3

Replace all occurrences of $u_{1}$ with $y$ .

$A \frac{y^{9} \cdot 0 - 1 \cdot 1 (9 y^{8} \frac{d}{d B} [y])}{{(y^{9})}^{2}} + \frac{d}{d B} [B e^{y}]$

Step 3.2.5

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

$A \frac{y^{9} \cdot 0 - 1 \cdot 1 (9 y^{8} y')}{{(y^{9})}^{2}} + \frac{d}{d B} [B e^{y}]$

Step 3.2.6

Multiply $y^{9}$ by $0$ .

$A \frac{0 - 1 \cdot 1 (9 y^{8} y')}{{(y^{9})}^{2}} + \frac{d}{d B} [B e^{y}]$

Step 3.2.7

Multiply $- 1$ by $1$ .

$A \frac{0 - (9 y^{8} y')}{{(y^{9})}^{2}} + \frac{d}{d B} [B e^{y}]$

Step 3.2.8

Multiply $9$ by $- 1$ .

$A \frac{0 - 9 (y^{8} y')}{{(y^{9})}^{2}} + \frac{d}{d B} [B e^{y}]$

Step 3.2.9

Subtract $9 y^{8} y'$ from $0$ .

$A \frac{- 9 y^{8} y'}{{(y^{9})}^{2}} + \frac{d}{d B} [B e^{y}]$

Step 3.2.10

Multiply the exponents in ${(y^{9})}^{2}$ .

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Step 3.2.10.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$A \frac{- 9 y^{8} y'}{y^{9 \cdot 2}} + \frac{d}{d B} [B e^{y}]$

Step 3.2.10.2

Multiply $9$ by $2$ .

$A \frac{- 9 y^{8} y'}{y^{18}} + \frac{d}{d B} [B e^{y}]$

Step 3.2.11

Cancel the common factor of $y^{8}$ and $y^{18}$ .

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Step 3.2.11.1

Factor $y^{8}$ out of $- 9 y^{8} y'$ .

$A \frac{y^{8} (- 9 y')}{y^{18}} + \frac{d}{d B} [B e^{y}]$

Step 3.2.11.2

Cancel the common factors.

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Step 3.2.11.2.1

Factor $y^{8}$ out of $y^{18}$ .

$A \frac{y^{8} (- 9 y')}{y^{8} y^{10}} + \frac{d}{d B} [B e^{y}]$

Step 3.2.11.2.2

Cancel the common factor.

$A \frac{y^{8} (- 9 y')}{y^{8} y^{10}} + \frac{d}{d B} [B e^{y}]$

Step 3.2.11.2.3

Rewrite the expression.

$A \frac{- 9 y'}{y^{10}} + \frac{d}{d B} [B e^{y}]$

Step 3.2.12

Move the negative in front of the fraction.

$A (- \frac{9 y'}{y^{10}}) + \frac{d}{d B} [B e^{y}]$

Step 3.2.13

Combine $A$ and $\frac{9 y'}{y^{10}}$ .

$- \frac{A (9 y')}{y^{10}} + \frac{d}{d B} [B e^{y}]$

Step 3.2.14

Move $9$ to the left of $A$ .

$- \frac{9 A y'}{y^{10}} + \frac{d}{d B} [B e^{y}]$

Step 3.3

Evaluate $\frac{d}{d B} [B e^{y}]$ .

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Step 3.3.1

Differentiate using the Product Rule which states that $\frac{d}{d B} [f (B) g (B)]$ is $f (B) \frac{d}{d B} [g (B)] + g (B) \frac{d}{d B} [f (B)]$ where $f (B) = B$ and $g (B) = e^{y}$ .

$- \frac{9 A y'}{y^{10}} + B \frac{d}{d B} [e^{y}] + e^{y} \frac{d}{d B} [B]$

Step 3.3.2

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

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Step 3.3.2.1

To apply the Chain Rule, set $u_{2}$ as $y$ .

$- \frac{9 A y'}{y^{10}} + B (\frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d B} [y]) + e^{y} \frac{d}{d B} [B]$

Step 3.3.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

$- \frac{9 A y'}{y^{10}} + B (e^{u_{2}} \frac{d}{d B} [y]) + e^{y} \frac{d}{d B} [B]$

Step 3.3.2.3

Replace all occurrences of $u_{2}$ with $y$ .

$- \frac{9 A y'}{y^{10}} + B (e^{y} \frac{d}{d B} [y]) + e^{y} \frac{d}{d B} [B]$

Step 3.3.3

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

$- \frac{9 A y'}{y^{10}} + B (e^{y} y') + e^{y} \frac{d}{d B} [B]$

Step 3.3.4

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

$- \frac{9 A y'}{y^{10}} + B (e^{y} y') + e^{y} \cdot 1$

Step 3.3.5

Multiply $e^{y}$ by $1$ .

$- \frac{9 A y'}{y^{10}} + B e^{y} y' + e^{y}$

Step 3.4

Simplify.

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Step 3.4.1

Reorder terms.

$e^{y} B y' - \frac{9 A y'}{y^{10}} + e^{y}$

Step 3.4.2

Reorder factors in $e^{y} B y' - \frac{9 A y'}{y^{10}} + e^{y}$ .

$B e^{y} y' - \frac{9 A y'}{y^{10}} + e^{y}$

Step 4

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

$0 = B e^{y} y' - \frac{9 A y'}{y^{10}} + e^{y}$

Step 5

Solve for $y'$ .

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Step 5.1

Rewrite the equation as $B e^{y} y' - \frac{9 A y'}{y^{10}} + e^{y} = 0$ .

$B e^{y} y' - \frac{9 A y'}{y^{10}} + e^{y} = 0$

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.

$1, y^{10}, 1, 1$

Step 5.2.2

The LCM of one and any expression is the expression.

$y^{10}$

Step 5.3

Multiply each term in $B e^{y} y' - \frac{9 A y'}{y^{10}} + e^{y} = 0$ by $y^{10}$ to eliminate the fractions.

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Step 5.3.1

Multiply each term in $B e^{y} y' - \frac{9 A y'}{y^{10}} + e^{y} = 0$ by $y^{10}$ .

$B e^{y} y' y^{10} - \frac{9 A y'}{y^{10}} y^{10} + e^{y} y^{10} = 0 y^{10}$

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $y^{10}$ .

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Step 5.3.2.1.1

Move the leading negative in $- \frac{9 A y'}{y^{10}}$ into the numerator.

$B e^{y} y' y^{10} + \frac{- 9 A y'}{y^{10}} y^{10} + e^{y} y^{10} = 0 y^{10}$

Step 5.3.2.1.2

Cancel the common factor.

$B e^{y} y' y^{10} + \frac{- 9 A y'}{y^{10}} y^{10} + e^{y} y^{10} = 0 y^{10}$

Step 5.3.2.1.3

Rewrite the expression.

$B e^{y} y' y^{10} - 9 A y' + e^{y} y^{10} = 0 y^{10}$

Step 5.3.2.2

Reorder factors in $B e^{y} y' y^{10} - 9 A y' + e^{y} y^{10}$ .

$B y' y^{10} e^{y} - 9 A y' + y^{10} e^{y} = 0 y^{10}$

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Multiply $0$ by $y^{10}$ .

$B y' y^{10} e^{y} - 9 A y' + y^{10} e^{y} = 0$

Step 5.4

Solve the equation.

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Step 5.4.1

Subtract $y^{10} e^{y}$ from both sides of the equation.

$B y' y^{10} e^{y} - 9 A y' = - y^{10} e^{y}$

Step 5.4.2

Factor $y'$ out of $B y' y^{10} e^{y} - 9 A y'$ .

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Step 5.4.2.1

Factor $y'$ out of $B y' y^{10} e^{y}$ .

$y' (B y^{10} e^{y}) - 9 A y' = - y^{10} e^{y}$

Step 5.4.2.2

Factor $y'$ out of $- 9 A y'$ .

$y' (B y^{10} e^{y}) + y' (- 9 A) = - y^{10} e^{y}$

Step 5.4.2.3

Factor $y'$ out of $y' (B y^{10} e^{y}) + y' (- 9 A)$ .

$y' (B y^{10} e^{y} - 9 A) = - y^{10} e^{y}$

Step 5.4.3

Divide each term in $y' (B y^{10} e^{y} - 9 A) = - y^{10} e^{y}$ by $B y^{10} e^{y} - 9 A$ and simplify.

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Step 5.4.3.1

Divide each term in $y' (B y^{10} e^{y} - 9 A) = - y^{10} e^{y}$ by $B y^{10} e^{y} - 9 A$ .

$\frac{y' (B y^{10} e^{y} - 9 A)}{B y^{10} e^{y} - 9 A} = \frac{- y^{10} e^{y}}{B y^{10} e^{y} - 9 A}$

Step 5.4.3.2

Simplify the left side.

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Step 5.4.3.2.1

Cancel the common factor of $B y^{10} e^{y} - 9 A$ .

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Step 5.4.3.2.1.1

Cancel the common factor.

$\frac{y' (B y^{10} e^{y} - 9 A)}{B y^{10} e^{y} - 9 A} = \frac{- y^{10} e^{y}}{B y^{10} e^{y} - 9 A}$

Step 5.4.3.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{- y^{10} e^{y}}{B y^{10} e^{y} - 9 A}$

Step 5.4.3.3

Simplify the right side.

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Step 5.4.3.3.1

Move the negative in front of the fraction.

$y' = - \frac{y^{10} e^{y}}{B y^{10} e^{y} - 9 A}$

Step 6

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

$\frac{d y}{d B} = - \frac{y^{10} e^{y}}{B y^{10} e^{y} - 9 A}$