Calculus Examples

$\frac{A}{y^{8}} + b e^{y} = z$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d z} (\frac{A}{y^{8}} + b e^{y}) = \frac{d}{d z} (z)$

Step 2

Differentiate the left side of the equation.

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Step 2.1

By the Sum Rule, the derivative of $\frac{A}{y^{8}} + b e^{y}$ with respect to $z$ is $\frac{d}{d z} [\frac{A}{y^{8}}] + \frac{d}{d z} [b e^{y}]$ .

$\frac{d}{d z} [\frac{A}{y^{8}}] + \frac{d}{d z} [b e^{y}]$

Step 2.2

Evaluate $\frac{d}{d z} [\frac{A}{y^{8}}]$ .

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Step 2.2.1

Since $\frac{1}{y^{8}}$ is constant with respect to $z$ , the derivative of $\frac{A}{y^{8}}$ with respect to $z$ is $\frac{1}{y^{8}} \frac{d}{d z} [A]$ .

$\frac{1}{y^{8}} \frac{d}{d z} [A] + \frac{d}{d z} [b e^{y}]$

Step 2.2.2

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

$\frac{1}{y^{8}} A' + \frac{d}{d z} [b e^{y}]$

Step 2.2.3

Combine $\frac{1}{y^{8}}$ and $A'$ .

$\frac{A'}{y^{8}} + \frac{d}{d z} [b e^{y}]$

Step 2.3

Differentiate using the Constant Rule.

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Step 2.3.1

Since $b e^{y}$ is constant with respect to $z$ , the derivative of $b e^{y}$ with respect to $z$ is $0$ .

$\frac{A'}{y^{8}} + 0$

Step 2.3.2

Add $\frac{A'}{y^{8}}$ and $0$ .

$\frac{A'}{y^{8}}$

Step 3

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

$1$

Step 4

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

$\frac{A'}{y^{8}} = 1$

Step 5

Solve for $A'$ .

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Step 5.1

Multiply both sides by $y^{8}$ .

$\frac{A'}{y^{8}} y^{8} = 1 y^{8}$

Step 5.2

Simplify.

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Step 5.2.1

Simplify the left side.

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Step 5.2.1.1

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

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Step 5.2.1.1.1

Cancel the common factor.

$\frac{A'}{y^{8}} y^{8} = 1 y^{8}$

Step 5.2.1.1.2

Rewrite the expression.

$A' = 1 y^{8}$

Step 5.2.2

Simplify the right side.

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Step 5.2.2.1

Multiply $y^{8}$ by $1$ .

$A' = y^{8}$

Step 6

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

$\frac{d A}{d z} = y^{8}$