Differentiate y = e^(2x) * artanh(sqrt(e^x - 1)) with respect to x.

Step-by-step solution

1. 1

   Establish the domain before differentiating. The square root needs $e^{x}-1\ge 0$, i.e. $x\ge 0$, and $\operatorname{artanh}(z)$ needs $|z|<1$, i.e. $e^{x}-1<1$, i.e. $x<\ln 2$. So the function only exists on

   $0\le x<\ln 2\approx 0.693$

   Knowing this explains the factor $2-e^{x}$ that appears later and why it never changes sign.

2. 2

   Set up the product rule. With $u=e^{2x}$ and $v=\operatorname{artanh}\left(\sqrt{e^{x}-1}\right)$,

   $y'=u' v+uv',\qquad u'=2e^{2x}$

   The hard half is $v'$, a two-layer composition.

3. 3

   Differentiate the inner square root. With $z=\left(e^{x}-1\right)^{1/2}$, the chain rule gives

   $z'=\frac12\left(e^{x}-1\right)^{-1/2}\cdot e^{x}=\frac{e^{x}}{2\sqrt{e^{x}-1}}$

4. 4

   Apply the artanh derivative and simplify $1-z^{2}$. Since $\dfrac{d}{dz}\operatorname{artanh}z=\dfrac{1}{1-z^{2}}$ and here $z^{2}=e^{x}-1$ exactly,

   $1-z^{2}=1-\left(e^{x}-1\right)=2-e^{x}$

   The square root disappears from this factor entirely — a genuine simplification, and the reason the answer stays readable. Note $2-e^{x}>0$ precisely on the domain found in step 1.

5. 5

   Assemble $v'$ and then $y'$.

   $v'=\frac{e^{x}}{2\sqrt{e^{x}-1}\left(2-e^{x}\right)}$

   $y'=2e^{2x}\operatorname{artanh}\left(\sqrt{e^{x}-1}\right)+\frac{e^{2x}\cdot e^{x}}{2\sqrt{e^{x}-1}\left(2-e^{x}\right)}=2e^{2x}\operatorname{artanh}\left(\sqrt{e^{x}-1}\right)+\frac{e^{3x}}{2\sqrt{e^{x}-1}\left(2-e^{x}\right)}$

6. 6

   Check numerically inside the domain. At $x=0.5$ a central difference of the original function gives $13.9758241$, and the formula gives $13.9758241$ ✓; at $x=0.6$ both give $28.7724865$ ✓. The derivative blows up at both ends of the domain, as the factors $\sqrt{e^{x}-1}\to 0$ and $2-e^{x}\to 0$ predict.