# e powers (cancel down)

• Oct 30th 2010, 09:04 AM
MattWT
e powers (cancel down)
Hello,

How do you cancel this equation down further?

Thanks

$((exp(x)exp(-x))/4) + ((exp(x)exp(-x))/4) + ((exp(-2x))/4) + ((exp(2x))/4)$
• Oct 30th 2010, 09:11 AM
Unknown008
$\dfrac{e^x e^{-x}}{4} + \dfrac{e^x e^{-x}}{4} + \dfrac{e^{-2x}}{4} + \dfrac{e^{2x}}{4}$

Is this what you mean?

Remember that:

$a^b a^c = a^{b+c}$
• Oct 30th 2010, 09:36 AM
Soroban
Hello, MattWT!

Are you working with hyperbolic functions?

Quote:

How do you cancel this equation down further?

. . $\dfrac{e^x\cdot e^{-x}}{4} + \dfrac{e^x\cdot e^{-x}}{4} + \dfrac{e^{-2x}}{4} + \dfrac{e^{2x}}{4}$

Since . $e^x\cdot e^{-x} \:=\:e^0 \:=\:1$

. . we have: . $\displaystyle \frac{1}{4} + \frac{1}{4} + \frac{e^{-2x}}{4} + \frac{e^{2x}}{4} \;=\;\frac{e^{2x}}{4} + \frac{1}{2} + \frac{e^{-2x}}{4}$

. . . . . . . . . . $\displaystyle =\;\frac{e^{2x} + 2 + e^{-2x}}{4} \;=\;\frac{(e^x + e^{-x})^2}{4}$

. . . . . . . . . . $=\;\left(\dfrac{e^x + e^{-x}}{2}\right)^2 \;=\;\cosh^2\!x$
• Oct 30th 2010, 09:42 AM
MattWT
I must have messed my algebra up at the beginning then, as the top line should be equal to 1:

The question was to dy/dx tanh(x) by writing the hyperbolic functions first.