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

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

Step 1

Differentiate both sides of the equation.

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

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.

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

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

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

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

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

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

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

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

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

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

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

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

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

$\frac{A'}{y^{8}} + e^{y} \cdot 1$

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

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

Step 4

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

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

Step 5

Solve for $A'$ .

Rewrite the equation as $\frac{A'}{y^{8}} + e^{y} = 0$ .

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

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

$\frac{A'}{y^{8}} = - e^{y}$

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

$\frac{A'}{y^{8}} y^{8} = - e^{y} y^{8}$

Simplify.

Simplify the left side.

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

Cancel the common factor.

$\frac{A'}{y^{8}} y^{8} = - e^{y} y^{8}$

Rewrite the expression.

$A' = - e^{y} y^{8}$

Simplify the right side.

Reorder factors in $- e^{y} y^{8}$ .

$A' = - y^{8} e^{y}$

Step 6

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

$\frac{d A}{d B} = - y^{8} e^{y}$

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

Related Questions

Support LCMGCF.com by disabling your adblocker.