Find the derivative of :
`f(x)=x+x^(2)+x^(3)+…..+x^(50)` at x = 1.

Answer

To find the derivative of the function \( f(x) = x + x^2 + x^3 + \ldots + x^{50} \) at \( x = 1 \), we can follow these steps:

### Step 1: Write down the function
The function is given as:
\[ f(x) = x + x^2 + x^3 + \ldots + x^{50} \]

### Step 2: Differentiate the function
We differentiate \( f(x) \) term by term. The derivative of \( x^n \) is \( n \cdot x^{n-1} \). Therefore, we have:
\[ f'(x) = \frac{d}{dx}(x) + \frac{d}{dx}(x^2) + \frac{d}{dx}(x^3) + \ldots + \frac{d}{dx}(x^{50}) \]
Calculating each term:
- The derivative of \( x \) is \( 1 \).
- The derivative of \( x^2 \) is \( 2x \).
- The derivative of \( x^3 \) is \( 3x^2 \).
- ...
- The derivative of \( x^{50} \) is \( 50x^{49} \).

Thus, we can express \( f'(x) \) as:
\[ f'(x) = 1 + 2x + 3x^2 + 4x^3 + \ldots + 50x^{49} \]

### Step 3: Evaluate the derivative at \( x = 1 \)
Now, we substitute \( x = 1 \) into the derivative:
\[ f'(1) = 1 + 2(1) + 3(1^2) + 4(1^3) + \ldots + 50(1^{49}) \]
This simplifies to:
\[ f'(1) = 1 + 2 + 3 + 4 + \ldots + 50 \]

### Step 4: Calculate the sum of the first 50 natural numbers
The sum of the first \( n \) natural numbers is given by the formula:
\[ \text{Sum} = \frac{n(n + 1)}{2} \]
For \( n = 50 \):
\[ \text{Sum} = \frac{50(50 + 1)}{2} = \frac{50 \times 51}{2} = \frac{2550}{2} = 1275 \]

### Final Result
Thus, the derivative of \( f(x) \) at \( x = 1 \) is:
\[ f'(1) = 1275 \]

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