Tangent line explorer

Pick a function and move the point $x_0$ with the slider or by dragging across the graph. The solid line is the tangent at that point. The dashed line is a secant through $x_0$ and a second point $x_0 + h$. Make $h$ small and the two lines become hard to tell apart.

Things to try

• Start with $x^2$ at $x_0 = 1$. Move $h$ toward the middle of its slider, first from the right and then from the left. The secant slope comes down to $2$ from above and climbs up to $2$ from below.
• Switch to $x^3 - 3x$ and find the two places where the tangent line is flat. Check them against $f'(x) = 3x^2 - 3$.
• With $e^x$, compare the height of the point with the tangent slope. They are the same number at every $x_0$.
• With $\ln x$, move $x_0$ toward $0$. The slope is $\frac{1}{x_0}$, so it doubles every time $x_0$ is cut in half.
• With $\sin x$, set $x_0 = 0$. The tangent there is the line $y = x$, which is why $\sin x \approx x$ when $x$ is small.
• With $\sqrt{x}$ at $x_0 = 2$, compare $h = 1$ with $h = -1$. The curve bends downward, so the secant to the right is flatter than the tangent and the secant to the left is steeper.

What you are seeing

The slope of the secant line is the difference quotient

The dotted right angle between the two points shows it as rise over run, with run $h$ and rise $f(x_0 + h) - f(x_0)$.

The derivative is the limit of that quotient as $h$ goes to $0$, when the limit exists.

The tangent line is the line through $(x_0, f(x_0))$ with slope $f'(x_0)$. The explorer writes it as $y = mx + b$ with $m = f'(x_0)$ and $b = f(x_0) - x_0 f'(x_0)$.

For $f(x) = x^2$ the quotient simplifies to $2x_0 + h$, so the secant slope misses the derivative $2x_0$ by exactly $h$. For the other functions the gap is not exactly $h$, but it still shrinks to $0$ as $h \to 0$.

The tangent slope comes from the exact derivative, such as $\frac{d}{dx}\sin x = \cos x$ or $\frac{d}{dx}\sqrt{x} = \frac{1}{2\sqrt{x}}$, not from a small $h$. For $\ln x$ and $\sqrt{x}$ the point stays at $x_0 > 0$. The slope of $\sqrt{x}$ grows without bound as $x_0 \to 0^+$, and there is no derivative at $0$ itself.

The rules behind these formulas are on the calculus reference sheet, and the videos are on the Calculus course page.