Find from first principles differential coefficients of ` sqrt( x^(2)+1) `

Answer

To find the differential coefficient of \( f(x) = \sqrt{x^2 + 1} \) from first principles, we will follow these steps:

### Step 1: Define the function
Let \( f(x) = \sqrt{x^2 + 1} \).

### Step 2: Apply the definition of the derivative
The derivative from first principles is defined as:
\[ f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} \]

### Step 3: Substitute \( f(x + \Delta x) \)
We need to calculate \( f(x + \Delta x) \):
\[ f(x + \Delta x) = \sqrt{(x + \Delta x)^2 + 1} \]
Now, substituting this into the derivative formula gives:
\[ f'(x) = \lim_{\Delta x \to 0} \frac{\sqrt{(x + \Delta x)^2 + 1} - \sqrt{x^2 + 1}}{\Delta x} \]

### Step 4: Simplify the expression
To simplify the expression, we can multiply the numerator and denominator by the conjugate of the numerator:
\[ f'(x) = \lim_{\Delta x \to 0} \frac{\left(\sqrt{(x + \Delta x)^2 + 1} - \sqrt{x^2 + 1}\right) \cdot \left(\sqrt{(x + \Delta x)^2 + 1} + \sqrt{x^2 + 1}\right)}{\Delta x \cdot \left(\sqrt{(x + \Delta x)^2 + 1} + \sqrt{x^2 + 1}\right)} \]
This results in:
\[ f'(x) = \lim_{\Delta x \to 0} \frac{(x + \Delta x)^2 + 1 - (x^2 + 1)}{\Delta x \cdot \left(\sqrt{(x + \Delta x)^2 + 1} + \sqrt{x^2 + 1}\right)} \]

### Step 5: Expand and simplify the numerator
Expanding the numerator:
\[ (x + \Delta x)^2 + 1 - (x^2 + 1) = (x^2 + 2x\Delta x + \Delta x^2 + 1) - (x^2 + 1) = 2x\Delta x + \Delta x^2 \]
Thus, we have:
\[ f'(x) = \lim_{\Delta x \to 0} \frac{2x\Delta x + \Delta x^2}{\Delta x \cdot \left(\sqrt{(x + \Delta x)^2 + 1} + \sqrt{x^2 + 1}\right)} \]

### Step 6: Factor out \( \Delta x \)
Factoring out \( \Delta x \) from the numerator:
\[ f'(x) = \lim_{\Delta x \to 0} \frac{\Delta x(2x + \Delta x)}{\Delta x \cdot \left(\sqrt{(x + \Delta x)^2 + 1} + \sqrt{x^2 + 1}\right)} \]
Cancelling \( \Delta x \) from numerator and denominator gives:
\[ f'(x) = \lim_{\Delta x \to 0} \frac{2x + \Delta x}{\sqrt{(x + \Delta x)^2 + 1} + \sqrt{x^2 + 1}} \]

### Step 7: Evaluate the limit
Now, as \( \Delta x \to 0 \):
\[ f'(x) = \frac{2x}{\sqrt{x^2 + 1} + \sqrt{x^2 + 1}} = \frac{2x}{2\sqrt{x^2 + 1}} = \frac{x}{\sqrt{x^2 + 1}} \]

### Final Answer
Thus, the differential coefficient of \( \sqrt{x^2 + 1} \) is:
\[ f'(x) = \frac{x}{\sqrt{x^2 + 1}} \]