Find y' for y = ln (e^x+e^-x+2)
Find $y'$ for $y = \ln\bigl(e^{x}+e^{-x}+2\bigr)$.
Explanation of Method
A logarithm whose derivative is a hyperbolic tangent.
Step 1: Apply the chain rule
With $u = e^{x}+e^{-x}+2$ and $u' = e^{x}-e^{-x}$: $y' = \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}+2}$.
Step 2: Factor both parts
The denominator is a perfect square, and the numerator a difference of squares, in the variable $e^{x/2}$: \[e^{x}+e^{-x}+2 = \left(e^{x/2}+e^{-x/2}\right)^{2}, \qquad e^{x}-e^{-x} = \left(e^{x/2}-e^{-x/2}\right)\left(e^{x/2}+e^{-x/2}\right).\]
Step 3: Cancel
\[ y' = \frac{\left(e^{x/2}-e^{-x/2}\right)\left(e^{x/2}+e^{-x/2}\right)}{\left(e^{x/2}+e^{-x/2}\right)^{2}} = \frac{e^{x/2}-e^{-x/2}}{e^{x/2}+e^{-x/2}} = \tanh\frac{x}{2}. \]
Step 4: Verification
\[x = 0:\quad y = \ln 4 \approx 1.38629,\quad y' = \tanh 0 = 0, \text{ the minimum of } y \;\;\checkmark\]
\[x = 2:\quad y' = \tanh 1 \approx 0.76159; \qquad \frac{y(2+10^{-5})-y(2-10^{-5})}{2\times 10^{-5}} \approx 0.76159 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = \dfrac{e^{x}-e^{-x}}{e^{x}+e^{-x}+2} = \tanh\dfrac{x}{2}\,}\]
Remark
Remark. The factorisation also simplifies $y$ itself: $y = 2\ln\!\left(e^{x/2}+e^{-x/2}\right) = 2\ln\!\left(2\cosh\tfrac{x}{2}\right)$. Differentiating that form gives $2\cdot\tfrac12\tanh\tfrac{x}{2}$, the same answer in one line. Since $|\tanh| \lt 1$, the function grows asymptotically like $|x|$.