Find y' for y = arcsin(tanh x)
Find $y'$ for $y = \arcsin(\tanh x)$.
Explanation of Method
A circular inverse function of a hyperbolic one — the Gudermannian.
Step 1: Apply the chain rule
With $u = \tanh x$ and $u' = \dfrac{1}{\cosh^{2}x}$: $y' = \frac{1}{\sqrt{1-\tanh^{2}x}}\cdot\frac{1}{\cosh^{2}x}$.
Step 2: Simplify the radical
Dividing the identity $\cosh^{2}x - \sinh^{2}x = 1$ by $\cosh^{2}x$ gives $1-\tanh^{2}x = \dfrac{1}{\cosh^{2}x}$, so
\[ \sqrt{1-\tanh^{2}x} = \frac{1}{\cosh x} \qquad (\text{since } \cosh x \gt 0). \]
Step 3: Combine
One factor of $\cosh x$ cancels: $y' = \cosh x\cdot\frac{1}{\cosh^{2}x} = \frac{1}{\cosh x}$.
Step 4: Verification
\[x = 0:\quad y = \arcsin 0 = 0,\quad y' = \frac{1}{\cosh 0} = 1 \;\;\checkmark\]
\[x = 1:\quad y' = \frac{1}{\cosh 1} \approx 0.64805; \qquad \frac{y(1+10^{-5})-y(1-10^{-5})}{2\times 10^{-5}} \approx 0.64805 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = \dfrac{1}{\cosh x} = \operatorname{sech} x\,}\]
Remark
Remark. This function is the Gudermannian, the bridge between circular and hyperbolic angles: it is defined for all real $x$ and increases from $-\tfrac{\pi}{2}$ to $\tfrac{\pi}{2}$. Read backwards, the result gives $\displaystyle\int\operatorname{sech} x\,dx = \arcsin(\tanh x)+C$.