If `y = (2(x-sinx)^(3/2))/(sqrtx)` then `(dy)/(dx)`

Answer

To find the derivative \(\frac{dy}{dx}\) for the function \(y = \frac{2(x - \sin x)^{3/2}}{\sqrt{x}}\), we will use the quotient rule and the chain rule of differentiation. Here’s a step-by-step solution:

### Step 1: Identify the function
We have:
\[ y = \frac{2(x - \sin x)^{3/2}}{\sqrt{x}} \]

### Step 2: Apply the Quotient Rule
The quotient rule states that if \(y = \frac{u}{v}\), then:
\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
where \(u = 2(x - \sin x)^{3/2}\) and \(v = \sqrt{x}\).

### Step 3: Differentiate \(u\) and \(v\)
1. **Differentiate \(u\)**:
\[ u = 2(x - \sin x)^{3/2} \]
Using the chain rule:
\[ \frac{du}{dx} = 2 \cdot \frac{3}{2}(x - \sin x)^{1/2} \cdot (1 - \cos x) = 3(x - \sin x)^{1/2}(1 - \cos x) \]

2. **Differentiate \(v\)**:
\[ v = \sqrt{x} = x^{1/2} \]
\[ \frac{dv}{dx} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}} \]

### Step 4: Substitute into the Quotient Rule
Now substitute \(u\), \(v\), \(\frac{du}{dx}\), and \(\frac{dv}{dx}\) into the quotient rule:
\[ \frac{dy}{dx} = \frac{\sqrt{x} \cdot 3(x - \sin x)^{1/2}(1 - \cos x) - 2(x - \sin x)^{3/2} \cdot \frac{1}{2\sqrt{x}}}{x} \]

### Step 5: Simplify the expression
1. Multiply out the terms:
\[ = \frac{3\sqrt{x}(x - \sin x)^{1/2}(1 - \cos x) - (x - \sin x)^{3/2} \cdot \frac{1}{\sqrt{x}}}{x} \]

2. Combine the terms:
\[ = \frac{3\sqrt{x}(x - \sin x)^{1/2}(1 - \cos x) - \frac{(x - \sin x)^{3/2}}{\sqrt{x}}}{x} \]

3. Factor out common terms if possible.

### Final Result
Thus, we have:
\[ \frac{dy}{dx} = \frac{3\sqrt{x}(x - \sin x)^{1/2}(1 - \cos x) - \frac{(x - \sin x)^{3/2}}{\sqrt{x}}}{x} \]