Find, from first principle, the derivative of the following w.r.t. x :
`logx^(2)`

Answer

To find the derivative of the function \( f(x) = \log(x^2) \) using the first principle of derivatives, we will follow these steps:

### Step 1: Define the function and the first principle of derivatives
The first principle of derivatives states that:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
For our function, we have:
\[ f(x) = \log(x^2) \]

### Step 2: Calculate \( f(x+h) \)
We need to find \( f(x+h) \):
\[ f(x+h) = \log((x+h)^2) \]

### Step 3: Substitute into the derivative formula
Now we substitute \( f(x+h) \) and \( f(x) \) into the derivative formula:
\[ f'(x) = \lim_{h \to 0} \frac{\log((x+h)^2) - \log(x^2)}{h} \]

### Step 4: Use the properties of logarithms
Using the property of logarithms \( \log(a) - \log(b) = \log\left(\frac{a}{b}\right) \), we can simplify:
\[ f'(x) = \lim_{h \to 0} \frac{\log\left(\frac{(x+h)^2}{x^2}\right)}{h} \]
This simplifies to:
\[ f'(x) = \lim_{h \to 0} \frac{\log\left(\frac{(x+h)^2}{x^2}\right)}{h} = \lim_{h \to 0} \frac{\log\left(\frac{(x+h)^2}{x^2}\right)}{h} \]

### Step 5: Simplify the logarithm
We can rewrite \( \frac{(x+h)^2}{x^2} \) as:
\[ \frac{(x+h)^2}{x^2} = \frac{x^2 + 2xh + h^2}{x^2} = 1 + \frac{2h}{x} + \frac{h^2}{x^2} \]
Thus, we have:
\[ f'(x) = \lim_{h \to 0} \frac{\log\left(1 + \frac{2h}{x} + \frac{h^2}{x^2}\right)}{h} \]

### Step 6: Use the limit property
As \( h \to 0 \), we can use the property \( \lim_{u \to 0} \frac{\log(1+u)}{u} = 1 \):
Let \( u = \frac{2h}{x} + \frac{h^2}{x^2} \). As \( h \to 0 \), \( u \to 0 \) as well. We can express \( h \) in terms of \( u \):
\[ h = \frac{xu}{2} \text{ (for small } h\text{)} \]
Thus:
\[ f'(x) = \lim_{u \to 0} \frac{\log(1+u)}{\frac{xu}{2}} = \frac{2}{x} \cdot 1 = \frac{2}{x} \]

### Final Result
Therefore, the derivative of \( f(x) = \log(x^2) \) is:
\[ f'(x) = \frac{2}{x} \]