Find, from first principle, the derivative of `sin^(-1)x/x` w.r.t. x.

Answer

To find the derivative of the function \( f(x) = \frac{\sin^{-1} x}{x} \) with respect to \( x \) using the first principle of derivatives, we will follow these steps:

### Step 1: Define the derivative using the first principle
The derivative of a function \( f(x) \) at a point \( x \) is defined as:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

### Step 2: Substitute \( f(x) \) into the formula
Here, we have:
\[ f(x) = \frac{\sin^{-1} x}{x} \]
So,
\[ f(x+h) = \frac{\sin^{-1}(x+h)}{x+h} \]
Now, substituting into the derivative formula:
\[ f'(x) = \lim_{h \to 0} \frac{\frac{\sin^{-1}(x+h)}{x+h} - \frac{\sin^{-1} x}{x}}{h} \]

### Step 3: Combine the fractions
To simplify the expression, we will combine the fractions in the numerator:
\[ f'(x) = \lim_{h \to 0} \frac{\sin^{-1}(x+h) \cdot x - \sin^{-1} x \cdot (x+h)}{h \cdot (x+h) \cdot x} \]
This gives us:
\[ f'(x) = \lim_{h \to 0} \frac{x \sin^{-1}(x+h) - (x+h) \sin^{-1} x}{h \cdot (x+h) \cdot x} \]

### Step 4: Analyze the limit
As \( h \to 0 \), both the numerator and denominator approach 0, leading to an indeterminate form \( \frac{0}{0} \). We can apply L'Hôpital's Rule, which states that if we have \( \frac{0}{0} \), we can take the derivative of the numerator and the derivative of the denominator.

### Step 5: Differentiate the numerator and denominator
1. Differentiate the numerator:
\[ \frac{d}{dh} \left( x \sin^{-1}(x+h) - (x+h) \sin^{-1} x \right) = x \cdot \frac{1}{\sqrt{1 - (x+h)^2}} - \sin^{-1} x \]
(The derivative of \( \sin^{-1}(x+h) \) is \( \frac{1}{\sqrt{1 - (x+h)^2}} \))

2. Differentiate the denominator:
\[ \frac{d}{dh} \left( h \cdot (x+h) \cdot x \right) = (x+h) \cdot x \]

### Step 6: Apply L'Hôpital's Rule
Now we apply L'Hôpital's Rule:
\[ f'(x) = \lim_{h \to 0} \frac{x \cdot \frac{1}{\sqrt{1 - (x+h)^2}} - \sin^{-1} x}{(x+h) \cdot x} \]

### Step 7: Substitute \( h = 0 \)
Substituting \( h = 0 \):
\[ f'(x) = \frac{x \cdot \frac{1}{\sqrt{1 - x^2}} - \sin^{-1} x}{x^2} \]

### Step 8: Simplify the expression
Thus, we have:
\[ f'(x) = \frac{1}{x \sqrt{1 - x^2}} - \frac{\sin^{-1} x}{x^2} \]

### Final Answer
The derivative of \( f(x) = \frac{\sin^{-1} x}{x} \) with respect to \( x \) is:
\[ f'(x) = \frac{1}{x \sqrt{1 - x^2}} - \frac{\sin^{-1} x}{x^2} \]

---