We want to show that Schwartz functions point-wise multiplied by $\tanh x$ are also Schwartz functions. To do this, we can need show that the nth derivative of $ \tanh ( x) \psi (x )$ decays sufficiently quickly. If we show all derivatives of $\tanh x$ are bounded, we are done, since the nth derivative of the function above is the binomial expansion of derivatives of $\tanh x$ and $\psi ( x)$. To do this, we prove that all derivatives of $\tanh x$ are polynomial functions of $\tanh x$. Since polynomials of bounded functions are themselves bounded functions, this shows that all derivatives of $\tanh x$ are bounded. So without further ado... CLAIM: For all $n \in \mathbb N$ , we can write the nth derivative of $\tanh x$ as a polynomial function of $\tanh x$. $ \frac { d ^ n } { d x ^ n} ( \tanh x ) = \Sigma _ { i \geq 0 } ^ N c_ i (\tanh x ) ^ i $for some sequence $c _ i$ where $c_i \in \mathbb R$, for some $N \in \mathbb N$. PROOF: We proceed by induction. For the case $n=1$, we have $\frac { d } { d x} ( \tanh x) = \text{sech} ^ 2 x = 1 - \tanh ^ 2 x $So, the first derivative is of the form above with $c_0 =0$ and $c_1 = - 1$. We assume that for all $k \leq n$ $ \frac { d ^ k } { d x ^k} ( \tanh x ) = \Sigma _ { i \geq 0 } ^ N c_ i (\tanh x ) ^ i $ For the case $k = n + 1$, we take the derivative. The constant term at $i=0$ vanishes. By the chain rule, and using our result for the first derivative, we have $ \frac{ d ^ { n + 1 }} { d x ^ { n + 1}} (\tanh x ) = \sum _ { i \geq 1 } ^ N c _ i \frac { d } { dx} \left [ ( \tanh x )^ i \right] = \sum _ { i \geq 1 } ^ N i \times c _ i \times ( 1 - \tanh ^ 2 x ) \times ( \tanh x ) ^ { i - 1 }$ But the right hand side is a polynomial function of a polynomial, which means that the (n+1)th derivative is also a polynomial. So we are done.