Derivative of arctan(2x): Answer, Proof, Mistakes

The proof: inverse-tangent rule plus chain rule

The standard fact is $\frac{d}{dx}\arctan u = \frac{u'}{1+u^2}$. Here $u = 2x$, so $u' = 2$ and $u^2 = 4x^2$.

The inner derivative $2$ lands in the numerator, and squaring the inner $2x$ produces $4x^2$ in the denominator. Both changes come from the same $u = 2x$.

Reading the graph of the slope

The denominator $1 + 4x^2$ is at least $1$ and grows without bound, so $\frac{2}{1+4x^2}$ is positive everywhere and never zero. That means $\arctan(2x)$ is always increasing.

The slope peaks at $x = 0$, where it equals $2$, and decays toward $0$ as $|x|$ grows. The curve flattens as it approaches its horizontal asymptotes $y = \pm\frac{\pi}{2}$.

Common mistakes

• Writing $\frac{2}{1+2x^2}$ by forgetting to square the whole inner function. It is $(2x)^2 = 4x^2$, not $2x^2$.
• Answering $\frac{1}{1+4x^2}$ and dropping the inner derivative $2$ that belongs in the numerator.
• Confusing it with $\frac{d}{dx}\arctan x = \frac{1}{1+x^2}$. The inner function $2x$ changes both the numerator and the denominator.
• Writing $\frac{2}{1+(2x)^2}$ but then expanding $(2x)^2$ incorrectly as $2x^2$ instead of $4x^2$.