Find, from first principle, the derivative of the following w.r.t. x :
`xe^(x)`

Answer

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

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

### Step 2: Substitute the function into the derivative formula
We have \( f(x) = x e^x \). Therefore,
\[ f(x + h) = (x + h)e^{x + h} \]
Now, substituting \( f(x + h) \) and \( f(x) \) into the derivative formula:
\[ f'(x) = \lim_{h \to 0} \frac{(x + h)e^{x + h} - x e^x}{h} \]

### Step 3: Expand \( f(x + h) \)
Using the property of exponents, we can rewrite \( e^{x + h} \) as \( e^x e^h \):
\[ f'(x) = \lim_{h \to 0} \frac{(x + h)e^x e^h - x e^x}{h} \]
Now, factor out \( e^x \):
\[ f'(x) = e^x \lim_{h \to 0} \frac{(x + h)e^h - x}{h} \]

### Step 4: Simplify the expression inside the limit
Now we simplify:
\[ f'(x) = e^x \lim_{h \to 0} \frac{xe^h + he^h - x}{h} \]
This can be separated into two terms:
\[ f'(x) = e^x \lim_{h \to 0} \left( \frac{xe^h - x}{h} + e^h \right) \]
The first term can be simplified:
\[ f'(x) = e^x \lim_{h \to 0} \left( x \frac{e^h - 1}{h} + e^h \right) \]

### Step 5: Evaluate the limits
As \( h \to 0 \), we know that \( \lim_{h \to 0} \frac{e^h - 1}{h} = 1 \) and \( \lim_{h \to 0} e^h = 1 \):
\[ f'(x) = e^x \left( x \cdot 1 + 1 \right) = e^x (x + 1) \]

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

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