Derivative graphs

What you are looking at

The height of the graph of $f'$ at any $x$ is the gradient of $f$ at that $x$. Here $f$, $f'$ and $f''$ are stacked on one $x$-axis, joined by a dashed vertical line through the point. As you drag the point along $f$, the tangent turns, and $f'$ is drawn in solid colour only as far as the point has travelled: the derivative graph is traced out by the slope you are looking at. The readout says whether $f$ is increasing, decreasing or stationary, and how it is concave.

With Max / min on, each turning point and inflection on $f$ is labelled with its coordinates, and the matching zeros of $f'$ and $f''$ are ringed. The point snaps onto these as you drag, so a horizontal tangent reads $f'(x) = 0$ exactly. Below the graph, the working gives $f'(x)$ and $f''(x)$ in symbols where it can, the tangent as $y = mx + c$, and, with Normal on, the equation of the normal with the check $m \times \left(-\dfrac{1}{m}\right) = -1$.

Try this

1. 1Start on $x^3 - 3x$ and drag the point from left to right: $f'$ is drawn in as you go and crosses zero exactly under the maximum $(-1, 2)$ and the minimum $(1, -2)$.
2. 2Drag through $x = 0$: $f''(0) = 0$ at the inflection $(0, 0)$, and the readout changes from concave down to concave up.
3. 3Switch to First principles and slide $h$ towards 0: the gradient of the red chord, $\dfrac{f(x+h)-f(x)}{h}$, closes in on $f'(x)$, shown beside it.
4. 4Turn on Normal and snap to the maximum: the tangent is flat, so the normal is the vertical line $x = -1$ and $-\dfrac{1}{m}$ has no value.

Mistakes students make

• Treating $f'(x) = 0$ as proof of a maximum or minimum. A stationary point can also be a point of inflection, as $x^3$ is at $x = 0$; check that $f'$ changes sign, or use the sign of $f''$.
• Assuming $f''(x) = 0$ means a point of inflection. For $x^4$, $f''(0) = 0$ but $x = 0$ is a minimum; the concavity must actually change, so $f''$ has to change sign.
• Confusing the zeros of $f$ with the zeros of $f'$. $f'$ is zero where $f$ is flat, not where it meets the $x$-axis: $x^3 - 3x$ has roots $0$ and $\pm\sqrt{3}$, but its stationary points are at $x = \pm 1$.

Where it is taught

• IB Maths AA SL/HL 5.2 · 5.4 · 5.7 · 5.8
• AP Calculus AB/BC Units 2 and 5
• Edexcel A-Level Pure Year 1 · Differentiation (tangents, normals, stationary points)
• Edexcel A-Level Pure Year 2 · Differentiation (second derivative, concavity, points of inflection)
• Greek Lykeio year 3 (Γ΄ Λυκείου) · §2.1 · §2.7