Find the differential coefficient of the following function with respect to 'x' `x/(sqrt(1-x^2))`

Answer

To find the differential coefficient of the function \( y = \frac{x}{\sqrt{1 - x^2}} \) with respect to \( x \), we can use the quotient rule of differentiation. The quotient rule states that if you have a function in the form \( \frac{u}{v} \), then the derivative \( \frac{dy}{dx} \) is given by:

\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

where \( u = x \) and \( v = \sqrt{1 - x^2} \).

### Step 1: Identify \( u \) and \( v \)
Let:
- \( u = x \)
- \( v = \sqrt{1 - x^2} \)

### Step 2: Compute \( \frac{du}{dx} \) and \( \frac{dv}{dx} \)
Now we differentiate \( u \) and \( v \):
- \( \frac{du}{dx} = 1 \)
- To differentiate \( v \), we use the chain rule:
\[ \frac{dv}{dx} = \frac{1}{2}(1 - x^2)^{-1/2} \cdot (-2x) = \frac{-x}{\sqrt{1 - x^2}} \]

### Step 3: Apply the Quotient Rule
Now we substitute \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \) into the quotient rule formula:
\[ \frac{dy}{dx} = \frac{\sqrt{1 - x^2} \cdot 1 - x \cdot \left(\frac{-x}{\sqrt{1 - x^2}}\right)}{(\sqrt{1 - x^2})^2} \]

### Step 4: Simplify the Expression
Now simplify the expression:
1. The numerator becomes:
\[ \sqrt{1 - x^2} + \frac{x^2}{\sqrt{1 - x^2}} = \frac{(1 - x^2) + x^2}{\sqrt{1 - x^2}} = \frac{1}{\sqrt{1 - x^2}} \]
2. The denominator is:
\[ (\sqrt{1 - x^2})^2 = 1 - x^2 \]

Putting it all together:
\[ \frac{dy}{dx} = \frac{\frac{1}{\sqrt{1 - x^2}}}{1 - x^2} = \frac{1}{(1 - x^2)^{3/2}} \]

### Final Answer
Thus, the differential coefficient of the function \( y = \frac{x}{\sqrt{1 - x^2}} \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \frac{1}{(1 - x^2)^{3/2}} \]