Find dA/dz z=A/(y^8)+Be^y

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

Step 1

Differentiate both sides of the equation.

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

Step 2

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

$1$

Step 3

Differentiate the right side of the equation.

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

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

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

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

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

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

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

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

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

Differentiate using the Constant Rule.

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

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

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

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

Step 4

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

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

Step 5

Solve for $A'$ .

Rewrite the equation as $\frac{A'}{y^{8}} = 1$ .

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

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

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

Simplify.

Simplify the left side.

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

Cancel the common factor.

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

Rewrite the expression.

$A' = 1 y^{8}$

Simplify the right side.

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

$A' = y^{8}$

Step 6

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

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

Find dA/dz z=A/(y^8)+Be^y

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