Derivative of x arctan x: Answer, Proof, Mistakes

The proof: product rule on x arctan x

Both factors depend on $x$, so the product rule applies: differentiate each factor in turn while holding the other fixed, then add.

Take $f(x) = x$ and $g(x) = \arctan x$, so $f'(x) = 1$ and $g'(x) = \frac{1}{1 + x^2}$ by the inverse trig rule of Topic 3.4.

Unlike $x\ln x$, where the second term collapses to 1, nothing simplifies here. An answer with one term in it is missing half the rule. Two units of the course meet in this one derivative: Topic 2.8 (The Product Rule) sets up the two terms, and Topic 3.4 (Differentiating Inverse Trigonometric Functions) supplies the second one.

What the derivative says about the graph

Both factors are odd, so $x\arctan x$ is even and symmetric about the $y$ axis. The derivative vanishes only at $x = 0$, where $\arctan 0 = 0$ and $\frac{0}{1} = 0$.

Differentiating a second time gives a tidy result.

That second derivative is positive for every $x$, so the graph is concave up everywhere and the critical point at the origin is the absolute minimum, with value 0.

The integration by parts mirror

The integral of $\arctan x$ is found by parts with $u = \arctan x$ and $dv = dx$, and the result is built out of this exact product.

Differentiate the right side: the first piece gives $\arctan x + \frac{x}{1 + x^2}$, and the correction term gives $-\frac{x}{1 + x^2}$, which cancels the second half and leaves $\arctan x$. The derivative and the integral are one statement read in opposite directions.

Common mistakes

• Answering $\frac{x}{1 + x^2}$ alone and dropping the $\arctan x$ term.
• Answering $\frac{1}{1 + x^2}$ by multiplying the two derivatives. A derivative of a product is never a product of derivatives.
• Using $\frac{1}{\sqrt{1 - x^2}}$, the $\arcsin$ derivative, in place of the $\arctan$ derivative.
• Cancelling $\frac{x}{1 + x^2}$ down to $\frac{1}{1 + x}$. The denominator is a sum, so the $x$ does not cancel.