Find y' for y = ln (2+e^-x+ sqrt(e^x+e^-x+4))

Find $y'$ for $y = \ln\!\left(2+e^{-x}+\sqrt{e^{x}+e^{-x}+4}\right)$.

Step 1: Introduce a shorthand

Let $s = \sqrt{e^{x}+e^{-x}+4}$, so that by the chain rule

\[ s' = \frac{e^{x}-e^{-x}}{2\sqrt{e^{x}+e^{-x}+4}} = \frac{e^{x}-e^{-x}}{2s}, \]

and set $u = 2+e^{-x}+s$, the argument of the logarithm.

Step 2: Differentiate the inner function

\[ u' = -e^{-x} + \frac{e^{x}-e^{-x}}{2s}. \]

Step 3: Form $y' = \dfrac{u'}{u}$ and clear the inner fraction

Multiplying numerator and denominator by $2s$: \[y' = \frac{-e^{-x} + \dfrac{e^{x}-e^{-x}}{2s}}{2+e^{-x}+s} = \frac{-2s\,e^{-x} + e^{x} - e^{-x}}{2s\bigl(2+e^{-x}+s\bigr)}.\]

Step 4: Group the numerator

Both negative terms carry the factor $e^{-x}$: $-2s\,e^{-x} + e^{x} - e^{-x} = e^{x} - e^{-x}\bigl(2s+1\bigr)$.

Step 5: Verification

\[x = 1:\quad e+e^{-1}+4 \approx 7.08616,\ s \approx 2.66199,\ u \approx 5.02986\]

\[y' \approx \frac{2.71828 - 0.36788(5.32398+1)}{2(2.66199)(5.02978)} \approx \frac{0.39182}{26.77884} \approx 0.014632\]

\[\frac{y(1+10^{-5})-y(1-10^{-5})}{2\times 10^{-5}} \approx 0.014632 \;\;\checkmark\]

Step 6: Why it does not simplify further

One might hope the radicand factors as in Problem 128, where $e^{x}+e^{-x}+2 = \bigl(e^{x/2}+e^{-x/2}\bigr)^{2}$. With $+4$ in place of $+2$ that fails: writing $t = e^{x/2}$,

\[ e^{x}+e^{-x}+4 = t^{2}+\frac{1}{t^{2}}+4 = \left(t+\frac1t\right)^{2}+2, \]

which is not a perfect square. So the answer is left as the single fraction above.