# How do you find the derivative of (3x)/(2e^x+e^-x)?

Nov 28, 2016

$= \frac{6 {e}^{x} \left(1 - x\right) + 3 {e}^{- x} \left(1 + x\right)}{2 {e}^{x} + {e}^{- x}} ^ 2$

#### Explanation:

Differentiating this function is determined by applying the
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quotient rule.
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$\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{u ' v - v ' u}{v} ^ 2$
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$\frac{d}{\mathrm{dx}} \left(\frac{3 x}{2 {e}^{x} + {e}^{- x}}\right)$
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$= \frac{\left(3 x\right) ' \times \left(2 {e}^{x} + {e}^{- x}\right) - \left(2 {e}^{x} + {e}^{- x}\right) ' \times \left(3 x\right)}{2 {e}^{x} + {e}^{- x}} ^ 2$
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$= \frac{3 \times \left(2 {e}^{x} + {e}^{- x}\right) - \left(2 {e}^{x} - {e}^{- x}\right) \times \left(3 x\right)}{2 {e}^{x} + {e}^{- x}} ^ 2$
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$= \frac{\left(6 {e}^{x} + 3 {e}^{- x}\right) - \left(6 x {e}^{x} - 3 x {e}^{- x}\right)}{2 {e}^{x} + {e}^{- x}} ^ 2$
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$= \frac{6 {e}^{x} + 3 {e}^{- x} - 6 x {e}^{x} + 3 x {e}^{- x}}{2 {e}^{x} + {e}^{- x}} ^ 2$
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$= \frac{6 {e}^{x} \left(1 - x\right) + 3 {e}^{- x} \left(1 + x\right)}{2 {e}^{x} + {e}^{- x}} ^ 2$