Find y' for y = e^xarctan x
Find $y'$ for $y = e^{x}\arctan x$.
Explanation of Method
A product of an exponential and an inverse trigonometric function.
Step 1: Identify the factors and apply the product rule
For $y = uv$ recall $y' = u'v + uv'$. Here
\[ u = e^{x},\quad u' = e^{x}, \qquad v = \arctan x,\quad v' = \frac{1}{1+x^{2}}. \]
Step 2: Substitute and factor
The common factor $e^{x}$ can be taken outside: \[y' = e^{x}\arctan x + e^{x}\cdot\frac{1}{1+x^{2}} = e^{x}\!\left(\arctan x + \frac{1}{1+x^{2}}\right).\]
Step 3: Verification
\[x = 0:\quad y' = 1\cdot(0+1) = 1; \qquad \frac{y(10^{-5})-y(-10^{-5})}{2\times 10^{-5}} \approx 1.0 \;\;\checkmark\]
\[x = 1:\quad y' = e\left(\frac{\pi}{4}+\frac12\right) \approx 3.494; \qquad \frac{y(1+10^{-5})-y(1-10^{-5})}{2\times 10^{-5}} \approx 3.494 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = e^{x}\!\left(\arctan x + \dfrac{1}{1+x^{2}}\right)\,}\]
Remark
Remark. The bracket vanishes where $\arctan x = -\dfrac{1}{1+x^{2}}$, which happens at a single negative value $x \approx -0.7472$ — the only critical point of $y$, and a minimum, since the bracket increases through zero there.