Find y' for y = (tanh x)/(coth x)

Find $y'$ for $y = \dfrac{\tanh x}{\coth x}$.

Step 1: Simplify the quotient

Since $\coth x = \dfrac{1}{\tanh x}$, dividing by $\coth x$ is the same as multiplying by $\tanh x$: \[y = \frac{\tanh x}{\coth x} = \tanh x\cdot\tanh x = \tanh^{2} x = \frac{\sinh^{2}x}{\cosh^{2}x}.\]

Step 2: Apply the product rule to $\tanh^{2}x$

Write $y = \tanh x\cdot\tanh x$ and use $\bigl(\tanh x\bigr)' = \dfrac{1}{\cosh^{2}x}$: \[y' = \frac{1}{\cosh^{2}x}\cdot\tanh x + \tanh x\cdot\frac{1}{\cosh^{2}x} = \frac{2\tanh x}{\cosh^{2}x}.\]

Step 3: Rewrite in terms of sinh and cosh

Substituting $\tanh x = \dfrac{\sinh x}{\cosh x}$: $y' = \frac{2}{\cosh^{2}x}\cdot\frac{\sinh x}{\cosh x} = \frac{2\sinh x}{\cosh^{3}x}$.

Step 4: Verification

\[\text{Chain rule:}\quad \bigl(\tanh^{2}x\bigr)' = 2\tanh x\cdot\frac{1}{\cosh^{2}x} \;\;\checkmark\]

\[x = 1:\quad y' = \frac{2\sinh 1}{\cosh^{3} 1} \approx \frac{2(1.1752)}{(1.5431)^{3}} \approx 0.6400 \;\;\checkmark\]

Step 5: A note on the domain

The simplified function $\tanh^{2}x$ is perfectly well behaved at $x = 0$, but the original expression is not: $\coth x = \dfrac{\cosh x}{\sinh x}$ is undefined at $x = 0$, where $\sinh 0 = 0$. The point $x = 0$ is therefore excluded from the domain of $y$, even though the limit

\[ \lim_{x\to 0}\frac{\tanh x}{\coth x} = \lim_{x\to 0}\tanh^{2}x = 0 \]

exists. This is a removable discontinuity: after filling in the value $0$ at the origin the function becomes $\tanh^{2}x$ on all of $\mathbb{R}$, and only then does it make sense to say that $x = 0$ is its minimum with $y'(0) = 0$.