If $f(x) = x^{100} + x^{99} + ... + x + 1$, then $f'(1)$ is equal to (A) $5050$ (B) $5049$ (C) $5051$ (D) $50051$

> [!TLDR] > To find $f'(1)$ for a polynomial, we first differentiate each term using the power rule, then substitute $x=1$. The problem simplifies to summing the first 100 natural numbers, which gives **5050**. The problem asks us to find the value of the derivative of a given polynomial function, $f(x)$, at a specific point, $x=1$. This involves two main steps: first, finding the general derivative $f'(x)$, and then evaluating this derivative at $x=1$. The concept of a derivative, $f'(x)$, represents the instantaneous rate of change of the function $f(x)$ with respect to $x$. Geometrically, it gives the slope of the tangent line to the curve $y=f(x)$ at any point $x$. For polynomials, differentiation is straightforward, relying on a few fundamental rules. 1. **Understand the function:** The given function is a sum of powers of $x$: $f(x) = x^{100} + x^{99} + \dots + x^2 + x^1 + 1$ We can write the constant term $1$ as $x^0$ to see a consistent pattern: $f(x) = x^{100} + x^{99} + \dots + x^2 + x^1 + x^0$. 2. **Recall differentiation rules for polynomials:** To find $f'(x)$, we use two primary rules: * **The Power Rule:** For any real number $n$, the derivative of $x^n$ with respect to $x$ is $nx^{n-1}$. > [!FORMULA] > $\frac{d}{dx}(x^n) = nx^{n-1}$ * **The Sum Rule:** The derivative of a sum of functions is the sum of their derivatives. If $f(x) = u(x) + v(x)$, then $f'(x) = u'(x) + v'(x)$. * **Derivative of a Constant:** The derivative of a constant term is $0$. This is a special case of the power rule where $n=0$, as $\frac{d}{dx}(c) = \frac{d}{dx}(c \cdot x^0) = c \cdot 0 \cdot x^{-1} = 0$. 3. **Differentiate $f(x)$ term by term:** Applying the power rule to each term in $f(x)$: * $\frac{d}{dx}(x^{100}) = 100x^{100-1} = 100x^{99}$ * $\frac{d}{dx}(x^{99}) = 99x^{99-1} = 99x^{98}$ * ... * $\frac{d}{dx}(x^2) = 2x^{2-1} = 2x^1 = 2x$ * $\frac{d}{dx}(x^1) = 1x^{1-1} = 1x^0 =

If f(x) = x¹⁰⁰ + x⁹⁹ + ... + x + 1, then f'(1) is equal to (A) 5050 (B) 5049 (C) 5051 (D) 50051

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f(x) = x^{100} + x^{99} + ... + x + 1

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Concept understanding — Derivative at a Point

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:

\frac{s(2+h) - s(2)}{h}

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 $t=2$ .

Note

This "shrinking interval" idea is the heart of the derivative. We're not setting $h=0$ (that would give $\frac{0}{0}$ , which is meaningless). We're letting $h$ approach 0 and seeing what the ratio approaches.

The Precise Definition

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

f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}

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.

f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}

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:

f'(3) = \lim_{h \to 0} \frac{(3+h)^2 - 3^2}{h}

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

f'(3) = \lim_{h \to 0} \frac{9 + 6h + h^2 - 9}{h} = \lim_{h \to 0} \frac{6h + h^2}{h}

Factor $h$ :

f'(3) = \lim_{h \to 0} \frac{h(6 + h)}{h} = \lim_{h \to 0} (6 + h)

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

Important

The derivative of $x^2$ at $x=3$ is 6. This means:

At $x=3$ , the function is increasing at a rate of 6 units per unit change in $x$ .
The tangent line to $y = x^2$ at $(3, 9)$ has slope 6.

Notation

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

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

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To find $f'(1)$ for a polynomial, we first differentiate each term using the power rule, then substitute $x=1$ . The problem simplifies to summing the first 100 natural numbers, which gives 5050.

The problem asks us to find the value of the derivative of a given polynomial function, $f(x)$ , at a specific point, $x=1$ . This involves two main steps: first, finding the general derivative $f'(x)$ , and then evaluating this derivative at $x=1$ .

The concept of a derivative, $f'(x)$ , represents the instantaneous rate of change of the function $f(x)$ with respect to $x$ . Geometrically, it gives the slope of the tangent line to the curve $y=f(x)$ at any point $x$ . For polynomials, differentiation is straightforward, relying on a few fundamental rules.

Understand the function:

The given function is a sum of powers of $x$ :

$f(x) = x^{100} + x^{99} + \dots + x^2 + x^1 + 1$

We can write the constant term $1$ as $x^0$ to see a consistent pattern:

$f(x) = x^{100} + x^{99} + \dots + x^2 + x^1 + x^0$ .
Recall differentiation rules for polynomials:

To find $f'(x)$ , we use two primary rules:
- The Power Rule: For any real number $n$ , the derivative of $x^n$ with respect to $x$ is $nx^{n-1}$ . $\frac{d}{dx}(x^n) = nx^{n-1}$
- The Sum Rule: The derivative of a sum of functions is the sum of their derivatives. If $f(x) = u(x) + v(x)$ , then $f'(x) = u'(x) + v'(x)$ .
- Derivative of a Constant: The derivative of a constant term is $0$ . This is a special case of the power rule where $n=0$ , as $\frac{d}{dx}(c) = \frac{d}{dx}(c \cdot x^0) = c \cdot 0 \cdot x^{-1} = 0$ .
Differentiate $f(x)$ term by term:

Applying the power rule to each term in $f(x)$ :
- $\frac{d}{dx}(x^{100}) = 100x^{100-1} = 100x^{99}$
- $\frac{d}{dx}(x^{99}) = 99x^{99-1} = 99x^{98}$
- ...
- $\frac{d}{dx}(x^2) = 2x^{2-1} = 2x^1 = 2x$
- $\frac{d}{dx}(x^1) = 1x^{1-1} = 1x^0 = 1$
- $\frac{d}{dx}(1) = \frac{d}{dx}(x^0) = 0x^{-1} = 0$ (or simply, the derivative of a constant is zero).
Formulate $f'(x)$ :

Combining these derivatives using the sum rule, we get:

$f'(x) = 100x^{99} + 99x^{98} + \dots + 2x + 1 + 0$

$f'(x) = 100x^{99} + 99x^{98} + \dots + 2x + 1$
Evaluate $f'(x)$ at $x=1$ :

Now, substitute $x=1$ into the expression for $f'(x)$ :

$f'(1) = 100(1)^{99} + 99(1)^{98} + \dots + 2(1) + 1$ …

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Q.If f(x)=x100+x99+...+x+1f(x) = x^{100} + x^{99} + ... + x + 1f(x)=x100+x99+...+x+1, then f′(1)f'(1)f′(1) is equal to (A) 505050505050 (B) 504950495049 (C) 505150515051 (D) 500515005150051

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Q.If $f(x) = x^{100} + x^{99} + ... + x + 1$ , then $f'(1)$ is equal to
(A) $5050$
(B) $5049$
(C) $5051$
(D) $50051$

Unlock everything free for 14 days