# Constant Multiple Rule

• Dec 5th 2009, 08:58 AM
Archduke01
Constant Multiple Rule
Finding the derivative of this rule;

y = (e^x - e^-x) / 2

y = 1/2 (e^x - e^-x)

y' = 1/2 (e^x + e^-x)

The constant multiple rule was applied to that last line to flip the minus sign to a plus ... but I don't understand the logic. How does that rule work?
• Dec 5th 2009, 09:10 AM
skeeter
Quote:

Originally Posted by Archduke01
Finding the derivative of this rule;

y = (e^x - e^-x) / 2

y = 1/2 (e^x - e^-x)

y' = 1/2 (e^x + e^-x)

The constant multiple rule was applied to that last line to flip the minus sign to a plus ... but I don't understand the logic. How does that rule work?

if $v = -e^u$ , then $v' = -e^u \cdot u'$

$v = -e^{-x}$ ... $v' = -e^{-x} \cdot (-1) = +e^{-x}$
• Dec 5th 2009, 09:13 AM
Haversine
Quote:

Originally Posted by Archduke01
Finding the derivative of this rule;

y = (e^x - e^-x) / 2

y = 1/2 (e^x - e^-x)

y' = 1/2 (e^x + e^-x)

The constant multiple rule was applied to that last line to flip the minus sign to a plus ... but I don't understand the logic. How does that rule work?

$\frac{d}{dx}y = \frac{d}{dx} \frac{1}{2} (e^x - e^{-x})$

$\frac{dy}{dx} = \frac{1}{2} \frac{d}{dx} (e^x - e^{-x})$

$\frac{dy}{dx} = \frac{1}{2} \left(\frac{d}{dx} e^x - \frac{d}{dx} e^{-x}\right)$

$\frac{dy}{dx} = \frac{1}{2} \left(\left(\frac{d}{dx} x\right)e^x - \left(\frac{d}{dx} (-x)\right) e^{-x}\right)$

$\frac{dy}{dx} = \frac{1}{2} \left((1)e^x - (-1) e^{-x}\right)$

$\frac{dy}{dx} = \frac{1}{2} (e^x + e^{-x})$