Derivative of arctan(e^x): Answer, Proof, Mistakes

Chain rule into the arctangent formula

Two facts do all the work here. The arctangent rule says $\frac{d}{dx}\arctan u = \frac{u'}{1+u^{2}}$, and the inner function $u = e^{x}$ is its own derivative.

The step that decides whether the answer is right is $\left(e^{x}\right)^{2} = e^{2x}$. A power raised to a power multiplies the exponents, so the $2$ lands on the outside of $x$ and gives $e^{2x}$. Writing $e^{x^{2}}$ instead squares the exponent, which is a different function entirely.

What the derivative says about the graph

The numerator $e^{x}$ is positive for every real $x$, and the denominator $1+e^{2x}$ is a positive number plus $1$. So the derivative is positive everywhere and $\arctan\left(e^{x}\right)$ is strictly increasing on the whole real line, with no critical points to hunt for.

It is also bounded. As $x \to -\infty$ the input $e^{x}$ shrinks to $0$ and the function approaches $0$; as $x \to \infty$ the input grows without bound and the function approaches $\frac{\pi}{2}$. The whole curve lives inside the strip $0 < y < \frac{\pi}{2}$ between two horizontal asymptotes.

Dividing the top and bottom by $e^{x}$ exposes a symmetry: swapping $x$ for $-x$ leaves the answer alone, so the derivative is an even function. Its peak is at $x = 0$, where it equals $\frac{1}{2}$, and that input is the inflection point of $\arctan\left(e^{x}\right)$. The sample values agree, and the symmetry shows up directly: the derivative is about $0.324$ at both $x = -1$ and $x = 1$, about $0.443$ at $x = 0.5$, and about $0.0497$ at $x = 3$.

The mistakes students make

Almost every wrong answer to this one comes from the denominator or from a missing factor of $e^{x}$.

• Answering $\frac{e^{x}}{1+e^{x^{2}}}$. The rule squares the inner function, and $\left(e^{x}\right)^{2}$ multiplies the exponents to give $e^{2x}$. Only $e^{x}$ composed with $x^{2}$ would produce $e^{x^{2}}$.
• Answering $\frac{1}{1+e^{2x}}$ by quoting the arctangent rule without the chain rule. That drops the factor $e^{x}$: at $x = 1$ it gives about $0.119$ instead of the correct $\frac{e}{1+e^{2}} \approx 0.324$.
• Cancelling the $e^{x}$ on top against the $e^{2x}$ below to get $\frac{1}{1+e^{x}}$. The denominator is a sum, not a product, so nothing cancels across the $+1$. At $x = 1$ that form gives $\frac{1}{1+e} \approx 0.269$ against the correct $0.324$; the two agree only at $x = 0$, where both equal $\frac{1}{2}$, so checking there proves nothing.