Find y' for y = ln (e^2x+1) - 2 arctan (e^x)
Find $y'$ for $y = \ln\bigl(e^{2x}+1\bigr) - 2\arctan\bigl(e^{x}\bigr)$.
Explanation of Method
A logarithm and an arctangent sharing the same denominator.
Step 1: Differentiate the logarithm
With $u = e^{2x}+1$ and $u' = 2e^{2x}$: $\frac{d}{dx}\ln\bigl(e^{2x}+1\bigr) = \frac{2e^{2x}}{e^{2x}+1}$.
Step 2: Differentiate the arctangent
With $u = e^{x}$, so that $1+u^{2} = 1+e^{2x}$: $\frac{d}{dx}\Bigl(-2\arctan e^{x}\Bigr) = -2\cdot\frac{e^{x}}{1+e^{2x}}$.
Step 3: Combine
The two denominators are identical, so the terms add directly: \[y' = \frac{2e^{2x}}{e^{2x}+1} - \frac{2e^{x}}{e^{2x}+1} = \frac{2e^{2x}-2e^{x}}{e^{2x}+1} = \frac{2e^{x}\bigl(e^{x}-1\bigr)}{e^{2x}+1}.\]
Step 4: Verification
\[x = 0:\quad y' = \frac{2\cdot 1\cdot 0}{2} = 0, \text{ a critical point} \;\;\checkmark\]
\[x = 1:\quad y' = \frac{2e(e-1)}{e^{2}+1} \approx \frac{2(2.71828)(1.71828)}{8.38906} \approx 1.11354 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = \dfrac{2e^{x}\bigl(e^{x}-1\bigr)}{e^{2x}+1}\,}\]
Remark
Remark. Since $e^{x} \gt 0$, the sign of $y'$ is the sign of $e^{x}-1$, i.e. of $x$: the function decreases on $(-\infty,0)$ and increases on $(0,\infty)$, with a global minimum at the origin.