Derivative of x/sqrt(x^2+1): Proof and Graph Shape

Quotient rule, then one clean power

The radical in the denominator is the only complication, and it behaves twice over. Differentiating $\sqrt{x^{2}+1}$ by the chain rule gives $\frac{x}{\sqrt{x^{2}+1}}$, while squaring it for the quotient rule denominator gives a plain $x^{2}+1$ with no radical left in it.

Multiply the top and bottom by $\sqrt{x^{2}+1}$ to clear the small fraction sitting inside the big one. The top becomes $\left(x^{2}+1\right)-x^{2}$, which is $1$.

Always increasing, always between two asymptotes

$\left(x^{2}+1\right)^{3/2}$ is positive for every real $x$, so $f'$ is never zero and never undefined. There are no critical points, no turning points and no places to test for a maximum: the function rises across the entire real line.

It rises without ever escaping the strip $-1 < f(x) < 1$, because $\sqrt{x^{2}+1}$ is always a little bigger than $\left|x\right|$. As $x\to\infty$ the ratio tends to $1$ and as $x\to-\infty$ it tends to $-1$, giving two horizontal asymptotes.

The derivative shows how a rising function stays bounded: it flattens out. At $x=0$ the slope is $1$, at $x=2$ it is $\frac{1}{5^{3/2}}\approx 0.089$, and it decays toward $0$ in both directions.

The mistakes students make

The first mistake lands on the wrong exponent, the second drops an inner derivative, and the third confuses a zero of f with a zero of f'.

• Combining the numerator over the radical to get $\frac{1}{\sqrt{x^{2}+1}}$ and then forgetting to divide by the quotient rule's outer denominator $x^{2}+1$. Clearing the inner fraction adds one power of $\sqrt{x^{2}+1}$ to that $x^{2}+1$, so the exponent is $\frac{3}{2}$, not $\frac{1}{2}$.
• Rewriting $f$ as $x\left(x^{2}+1\right)^{-1/2}$ and then differentiating the second factor as $-\frac{1}{2}\left(x^{2}+1\right)^{-3/2}$, dropping the inner $2x$. The correct piece is $-x\left(x^{2}+1\right)^{-3/2}$, and it is that $x$ that lets the two product rule terms combine into a single $1$.
• Calling $x=0$ a critical point because $f(0)=0$. A zero of the function is not a zero of the derivative: $f'(0)=1$, the steepest slope on the graph.