Q.For the function $f(x) = \dfrac{x^{100}}{100} + \dfrac{x^{99}}{99} + \cdots + \dfrac{x^2}{2} + x + 1$. Prove that $f'(1) = 100\,f'(0)$.

Derivative at a Point: From Intuition to Precision

Imagine you're driving a car. Your speedometer doesn't tell you your average speed over the whole trip — it tells you your speed right now, at this exact instant. That's the core idea of a derivative at a point: it measures how fast something is changing at a single moment.

The Intuition: Instantaneous Rate of Change

Let's start with something simpler. Suppose you drop a ball from a height. The distance it has fallen after $t$ seconds is given by $s(t) = 4.9t^2$ metres (ignoring air resistance).

If I ask you "how fast was the ball falling after exactly 2 seconds?", you can't just divide distance by time — that gives an average speed over an interval. You need the speed at $t=2$, not between $t=1$ and $t=3$.

Here's the trick: take a very small time interval around $t=2$, say from $t=2$ to $t=2+h$ where $h$ is tiny. The average speed over that interval is:

If $h = 0.1$, you get one number. If $h = 0.01$, you get a slightly different number. As $h$ gets closer and closer to 0, these average speeds approach a single value — that's the instantaneous speed at $t=2$.

The Precise Definition

For a function $f(x)$, the derivative at a point $x = a$ is defined as:

provided this limit exists.

Let's break this down piece by piece:

• $f(a+h) - f(a)$ is the change in the function's value when you move from $a$ to $a+h$.
• Dividing by $h$ gives the average rate of change over that interval.
• Taking the limit as $h \to 0$ shrinks the interval to a single point, giving the instantaneous rate of change.

Geometric Interpretation

There's also a beautiful geometric meaning. The average rate of change $\frac{f(a+h) - f(a)}{h}$ is the slope of the secant line through the points $(a, f(a))$ and $(a+h, f(a+h))$.

As $h \to 0$, these two points get closer together, and the secant line approaches a line that just touches the curve at $x=a$ — the tangent line. So:

The derivative at a point equals the slope of the tangent line to the curve at that point.

A Concrete Example

Let's compute the derivative of $f(x) = x^2$ at $x = 3$.

Using the definition:

Expand $(3+h)^2 = 9 + 6h + h^2$:

Factor $h$:

Since $h \to 0$, this approaches $6$.

Notation

You'll see several notations for "the derivative of $f$ at $x=a$":

• $f'(a)$ — Lagrange notation (most common) …