# Thread: tanh differentiable

1. ## tanh differentiable

$\displaystyle tanh(x)= \frac{e^x- e^{-x}}{e^x+ e^{-x}}$
(a) Prove that tanh is differentiable.

exp is differentiable, therefore tanh is differentiable?? The above has substraction and division by exp. How do i explain that tanh is still differentiable.

(b) Prove that tanh is strictly increasing, and hence invertible

the derivative of tanh is greater than zero, therefore tanh is strictly increasing. How do i prove its invertible.

Many thanks for any help.

2. Originally Posted by charikaar
How do I prove that tanh is differentiable?
By differentiating it.

3. Originally Posted by charikaar
$\displaystyle tanh(x)= \frac{e^x- e^{-x}}{e^x+ e^{-x}}$
(a) Prove that tanh is differentiable.

exp is differentiable, therefore tanh is differentiable?? The above has substraction and division by exp. How do i explain that tanh is still differentiable.
$\displaystyle \frac{d}{dx} [tanh(x)] = \frac{d}{dx} \frac{sinh(x)}{cosh(x)}$ $\displaystyle =\frac{cosh(x) \frac{d}{dx} [sinh(x)]-sinh(x)\frac{d}{dx}[cosh(x)]}{cosh^2 (x)}$

$\displaystyle = \frac{cosh^2(x)-sinh^2(x)}{cosh^2(x)}=\frac{1}{cosh^2(x)} = sech^2(x)$

You see it's easy to prove that it is differentiable, isn't it? You just need to know the approperiate identities.