Find dB/dA z=A/(y^14)+Be^y

$z = \frac{A}{y^{14}} + B e^{y}$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d A} (z) = \frac{d}{d A} (\frac{A}{y^{14}} + B e^{y})$

Step 2

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

$0$

Step 3

Differentiate the right side of the equation.

By the Sum Rule, the derivative of $\frac{A}{y^{14}} + B e^{y}$ with respect to $A$ is $\frac{d}{d A} [\frac{A}{y^{14}}] + \frac{d}{d A} [B e^{y}]$ .

$\frac{d}{d A} [\frac{A}{y^{14}}] + \frac{d}{d A} [B e^{y}]$

Evaluate $\frac{d}{d A} [\frac{A}{y^{14}}]$ .

Since $\frac{1}{y^{14}}$ is constant with respect to $A$ , the derivative of $\frac{A}{y^{14}}$ with respect to $A$ is $\frac{1}{y^{14}} \frac{d}{d A} [A]$ .

$\frac{1}{y^{14}} \frac{d}{d A} [A] + \frac{d}{d A} [B e^{y}]$

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

$\frac{1}{y^{14}} \cdot 1 + \frac{d}{d A} [B e^{y}]$

Multiply $\frac{1}{y^{14}}$ by $1$ .

$\frac{1}{y^{14}} + \frac{d}{d A} [B e^{y}]$

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

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

$\frac{1}{y^{14}} + e^{y} \frac{d}{d A} [B]$

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

$\frac{1}{y^{14}} + e^{y} B'$

Reorder terms.

$e^{y} B' + \frac{1}{y^{14}}$

Step 4

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

$0 = e^{y} B' + \frac{1}{y^{14}}$

Step 5

Solve for $B'$ .

Rewrite the equation as $e^{y} B' + \frac{1}{y^{14}} = 0$ .

$e^{y} B' + \frac{1}{y^{14}} = 0$

Reorder factors in $e^{y} B' + \frac{1}{y^{14}}$ .

$B' e^{y} + \frac{1}{y^{14}} = 0$

Subtract $\frac{1}{y^{14}}$ from both sides of the equation.

$B' e^{y} = - \frac{1}{y^{14}}$

Divide each term in $B' e^{y} = - \frac{1}{y^{14}}$ by $e^{y}$ and simplify.

Divide each term in $B' e^{y} = - \frac{1}{y^{14}}$ by $e^{y}$ .

$\frac{B' e^{y}}{e^{y}} = \frac{- \frac{1}{y^{14}}}{e^{y}}$

Simplify the left side.

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

Cancel the common factor.

$\frac{B' e^{y}}{e^{y}} = \frac{- \frac{1}{y^{14}}}{e^{y}}$

Divide $B'$ by $1$ .

$B' = \frac{- \frac{1}{y^{14}}}{e^{y}}$

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$B' = - \frac{1}{y^{14}} \cdot \frac{1}{e^{y}}$

Multiply $\frac{1}{e^{y}}$ by $\frac{1}{y^{14}}$ .

$B' = - \frac{1}{e^{y} y^{14}}$

Reorder factors in $- \frac{1}{e^{y} y^{14}}$ .

$B' = - \frac{1}{y^{14} e^{y}}$

Step 6

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

$\frac{d B}{d A} = - \frac{1}{y^{14} e^{y}}$

Find dB/dA z=A/(y^14)+Be^y

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