Differentiate $g(t)= {e^t - e^{-t} \over e^t + e^{-t}}$ I'm having some trouble trying to differentiate the function $g(t)= \dfrac{e^t - e^{-t}}{e^t + e^{-t}}$
Can someone help me? Thanks a lot! 
 A: Note that $g(t)=1$ for all $t$ for which it’s defined, i.e., for all $t\ne 0$.
Note: This answer was for the original version, which had
$$g(t)=\frac{e^t-e^{-t}}{e^t-e^{-t}}\;.$$
The corrected version can be differentiated as it stands by application of the quotient rule, or it can be manipulated a bit first to make the differentiation a little simpler:
$$g(t)=\frac{e^t-e^{-t}}{e^t+e^{-t}}\cdot\frac{e^t}{e^t}=\frac{e^{2t}-1}{e^{2t}+1}=1-\frac2{e^{2t}+1}=1-2\left(e^{2t}+1\right)^{-1}\;,$$
so
$$g'(t)=(-2)(-1)\left(e^{2t}+1\right)^{-2}\left(2e^{2t}\right)=\frac{4e^{2t}}{\left(e^{2t}+1\right)^2}\;.$$
A: You can cancel the terms in the numerator and denominator: the numerator is exactly the same function as is the denominator.
EDIT:  Okay...now it makes more sense! ;-)
Recall the quotient rule: $$g(t) = \frac{f(t)}{h(t)}$$
$$g'(t) = \frac{f'(t)h(t) - h'(t)f(t)}{[h(t)]^2}\tag{1}$$
In your case, $$g(t) = \frac{e^t - e^{-t}}{e^t - e^{-t}} = \frac{f(t)}{h(t)}$$
$f'(t) = e^t + e^{-t},\; h'(t) = e^t - e^{-t}$, so using the quotient rule $(1)$, find $g'(t)$ and simplify.
A: $g(t) = \dfrac{a(t)}{b(t)}$
$g'(t)= \dfrac{a'(t)b(t)-a(t)b'(t)}{b^2(t)}$
$g'(t)= \dfrac{(e^t+e^{-t})(e^t+e^{-t})-(e^t-e^{-t})(e^t-e^{-t})}{(e^t+e^{-t})^2}$
$g'(t)= \dfrac{(e^{2t}+2+e^{-2t})-(e^{2t}-2+e^{-2t})}{(e^t+e^{-t})^2} = \dfrac{4}{(e^t+e^{-t})^2}$

By the way,  $g(t)=\tanh(t)$, and   $g'(t)=\dfrac{1}{\cosh^2(t)} = \dfrac{1}{\left(\dfrac{e^t+e^{-t}}{2}\right)^2} = \dfrac{4}{(e^t+e^{-t})^2}$.
A: Note that $g(t) = \frac{\sinh(t)}{\cosh(t)}$. You can use quotient rule and remember $\sinh'(t) = \cosh(t), \cosh'(t) = \sinh(t), \cosh^2(t) - \sinh^2(t) = 1$.
