Q.Find the points on the curve $y = 4x^3 - 2x^5$, at which the tangent passes through the origin.

Tangent Line Slope: From Intuition to Precision

Imagine you're cycling up a winding hill road. At every point on that road, your front wheel points in a specific direction — that direction at that exact spot is what the tangent line captures. The slope of that tangent line is simply the steepness of the hill right under your wheel, not the average steepness over the last kilometre.

The Intuitive Idea

Take a curve — say, the graph of $y = x^2$, a simple upward-opening parabola. Pick a point on it, like $(1, 1)$. If you zoom in very close around that point, the curve starts to look almost straight. That nearly-straight line you see is the tangent line at that point. Its slope tells you: "If I move a tiny step to the right from here, how much does the curve go up or down?"

For a straight line, the slope is constant — the line is its own tangent everywhere. For a curve, the slope changes from point to point. That's the whole game.

The Precise Definition

We need to turn "zoom in very close" into mathematics. Here's how.

Take a curve $y = f(x)$ and a point $P = (a, f(a))$ on it. To find the slope of the tangent at $P$, we first consider a second point $Q = (a+h, f(a+h))$ nearby, where $h$ is a small number (positive or negative). The line through $P$ and $Q$ is a secant line. Its slope is:

Now, as $h$ gets smaller and smaller — as $Q$ slides along the curve toward $P$ — the secant line pivots and approaches a limiting position. That limiting line is the tangent line. Its slope is the limit of the secant slopes as $h$ approaches 0:

This limit, when it exists, is called the derivative of $f$ at $a$, denoted $f'(a)$.

A Concrete Example

For $f(x) = x^2$ at $x = 1$:

$f'(1) = \lim_{h \to 0} \frac{(1+h)^2 - 1^2}{h} = \lim_{h \to 0} \frac{1 + 2h + h^2 - 1}{h} = \lim_{h \to 0} \frac{2h + h^2}{h} = \lim_{h \to 0} (2 + h) = 2$ …