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

Jul 31, 2016

$\frac{\mathrm{dY}}{\mathrm{dx}} = \frac{4}{{e}^{x} + {e}^{-} x} ^ 2$.

#### Explanation:

To find the derivative, we have to use the Quotient Rule, and, the Chain Rule, given below for ready reference :-

The Quotient Rule :- $\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$.

By the Chain Rule, $\frac{d}{\mathrm{dx}} {e}^{a x} = {e}^{a x} \cdot \frac{d}{\mathrm{dx}} \left(a x\right) = a \cdot {e}^{a x}$.

As a particular case of this, we have, $\frac{d}{\mathrm{dx}} {e}^{-} x = - {e}^{-} x$.

Hence,

$\frac{\mathrm{dY}}{\mathrm{dx}}$

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

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

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

$= \frac{4}{{e}^{x} + {e}^{-} x} ^ 2$.

In fact, if we use hyperbolic funs., then, since $Y = \tanh x$, we can directly say that $\frac{\mathrm{dY}}{\mathrm{dx}} = {\sech}^{2} x = \frac{4}{{e}^{x} + {e}^{-} x} ^ 2$.