Slope at a point versus the derivative function
The derivative \(f'(x)\) is a function: feed it an \(x\) and it returns a number. That number is the slope of the curve at that point, and it is what "the derivative at a point" means. Evaluating it is a two-step job — differentiate first, then substitute.
The order matters. Substituting before differentiating turns \(f(x)\) into a constant, whose derivative is zero. This is one of the most common errors in a first calculus course, and it always produces the same wrong answer: 0.
A worked example, end to end
Take \(f(x) = x^{3} - 3x\) at \(a = 2\).
- Differentiate: \(f'(x) = 3x^{2} - 3\).
- Substitute: \(f'(2) = 3(4) - 3 = 9\). That is the slope.
- Find the point: \(f(2) = 8 - 6 = 2\), so the curve passes through \((2, 2)\).
- Assemble: \(y - 2 = 9(x - 2)\), which tidies to \(y = 9x - 16\).
Set the point control to 2 above and every one of those numbers appears. Move it to 1 and the slope becomes 0 — the curve has a stationary point there, and the tangent lies flat.
Finding the tangent line equation
Once you have the slope \(m = f'(a)\) and the point \((a, f(a))\), the tangent line follows from point-slope form:
\[y - f(a) = f'(a)\,(x - a)\]
Rearranged into \(y = mx + b\), the intercept is \(b = f(a) - f'(a)\,a\). The calculator above prints the finished equation and draws the line on the curve, so you can check that it touches at the right place and leans at the right angle.
The normal line
The normal is perpendicular to the tangent at the same point, so its slope is \(-1/f'(a)\), provided \(f'(a) \neq 0\). Where the tangent is horizontal the normal is vertical, and has no slope at all.
When the derivative does not exist there
A derivative at a point can fail to exist even when the function is perfectly well defined there:
- Corners. \(f(x) = |x|\) at \(x = 0\): the slope approaches \(-1\) from the left and \(+1\) from the right, so there is no single value.
- Vertical tangents. \(f(x) = \sqrt[3]{x}\) at \(x = 0\): the slope runs off to infinity.
- Discontinuities. If \(f\) jumps at \(a\), no tangent can exist there. Differentiability implies continuity, never the other way round.
Try abs(x) at 0 in the calculator: the derivative it reports uses \(\operatorname{sgn}(x)\), which is correct on either side and undefined exactly at the corner.
Common mistakes
- Substituting before differentiating. \(f(2) = 2\) is a constant; differentiating it gives 0, not 9.
- Reporting the point value \(f(a)\) when the question asked for the slope \(f'(a)\).
- Writing the tangent line through the origin. It must pass through \((a, f(a))\), not \((0, 0)\).
- Assuming a slope exists everywhere. Corners and vertical tangents are common in exam questions precisely because they catch this.
Questions
How do I find the derivative at a specific point?
Differentiate the function first, then substitute the point into the result. The calculator above does both — enter the function and set the point control.
What is the difference between f(a) and f′(a)?
\(f(a)\) is the height of the curve at \(a\); \(f'(a)\) is its steepness there. A question asking for "the value" wants the first, and "the slope", "the rate of change" or "the gradient" wants the second.
How do I find the equation of the tangent line?
Use \(y - f(a) = f'(a)(x - a)\). You need two ingredients: the point \((a, f(a))\) and the slope \(f'(a)\).
What does it mean when the derivative at a point is zero?
The tangent is horizontal — a maximum, a minimum, or a flat inflection. Which one it is depends on the second derivative or on how the sign of \(f'\) changes around it.
Can the derivative fail to exist at a point?
Yes — at corners such as \(|x|\) at zero, at vertical tangents such as \(\sqrt[3]{x}\) at zero, and anywhere the function is discontinuous.