Find the derivative of `f(x)=x^(n)`, where n is positive integer, by first principle.

Answer

To find the derivative of the function \( f(x) = x^n \) using the first principle of derivatives, we will follow these steps:

### Step 1: Write the definition of the derivative using the first principle.
The derivative of \( f(x) \) at a point \( x \) is given by:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

### Step 2: Substitute \( f(x) \) into the formula.
Given \( f(x) = x^n \), we need to find \( f(x+h) \):
\[ f(x+h) = (x+h)^n \]
Now, substituting into the derivative formula, we have:
\[ f'(x) = \lim_{h \to 0} \frac{(x+h)^n - x^n}{h} \]

### Step 3: Expand \( (x+h)^n \) using the Binomial Theorem.
According to the Binomial Theorem:
\[ (x+h)^n = x^n + \binom{n}{1} x^{n-1} h + \binom{n}{2} x^{n-2} h^2 + \ldots + h^n \]
This simplifies to:
\[ (x+h)^n = x^n + n x^{n-1} h + \frac{n(n-1)}{2} x^{n-2} h^2 + \ldots + h^n \]

### Step 4: Substitute the expansion back into the limit.
Now substituting this expansion back into our derivative expression:
\[ f'(x) = \lim_{h \to 0} \frac{\left(x^n + n x^{n-1} h + \frac{n(n-1)}{2} x^{n-2} h^2 + \ldots + h^n\right) - x^n}{h} \]
This simplifies to:
\[ f'(x) = \lim_{h \to 0} \frac{n x^{n-1} h + \frac{n(n-1)}{2} x^{n-2} h^2 + \ldots + h^n}{h} \]

### Step 5: Simplify the expression.
Dividing each term by \( h \):
\[ f'(x) = \lim_{h \to 0} \left(n x^{n-1} + \frac{n(n-1)}{2} x^{n-2} h + \ldots + h^{n-1}\right) \]

### Step 6: Evaluate the limit as \( h \to 0 \).
As \( h \) approaches 0, all terms involving \( h \) vanish:
\[ f'(x) = n x^{n-1} \]

### Conclusion:
Thus, the derivative of \( f(x) = x^n \) is:
\[ f'(x) = n x^{n-1} \]