Differentiate `sin^(-1)(4xsqrt(1-4x^(2)))w.r.t.sqrt(1-4x^(2))`, if `x in(-(1)/(2sqrt2),(1)/(2sqrt2))`

Answer

To differentiate \( \sin^{-1}(4x\sqrt{1-4x^2}) \) with respect to \( \sqrt{1-4x^2} \), we will follow these steps:

### Step 1: Define the Functions
Let:
- \( y_1 = \sin^{-1}(4x\sqrt{1-4x^2}) \)
- \( y_2 = \sqrt{1-4x^2} \)

### Step 2: Use the Chain Rule
We want to find \( \frac{dy_1}{dy_2} \). By the chain rule:
\[ \frac{dy_1}{dy_2} = \frac{dy_1}{dx} \cdot \frac{dx}{dy_2} \]

### Step 3: Differentiate \( y_1 \) with respect to \( x \)
Using the derivative of the inverse sine function:
\[ \frac{dy_1}{dx} = \frac{1}{\sqrt{1 - (4x\sqrt{1-4x^2})^2}} \cdot \frac{d}{dx}(4x\sqrt{1-4x^2}) \]
Now, we need to differentiate \( 4x\sqrt{1-4x^2} \):
Using the product rule:
\[ \frac{d}{dx}(4x\sqrt{1-4x^2}) = 4\sqrt{1-4x^2} + 4x \cdot \frac{d}{dx}(\sqrt{1-4x^2}) \]
Now, differentiate \( \sqrt{1-4x^2} \):
\[ \frac{d}{dx}(\sqrt{1-4x^2}) = \frac{1}{2\sqrt{1-4x^2}} \cdot (-8x) = -\frac{4x}{\sqrt{1-4x^2}} \]
Substituting this back:
\[ \frac{d}{dx}(4x\sqrt{1-4x^2}) = 4\sqrt{1-4x^2} - \frac{16x^2}{\sqrt{1-4x^2}} = \frac{4(1-4x^2) - 16x^2}{\sqrt{1-4x^2}} = \frac{4 - 32x^2}{\sqrt{1-4x^2}} \]
Thus, we have:
\[ \frac{dy_1}{dx} = \frac{1}{\sqrt{1 - (4x\sqrt{1-4x^2})^2}} \cdot \frac{4 - 32x^2}{\sqrt{1-4x^2}} \]

### Step 4: Differentiate \( y_2 \) with respect to \( x \)
Now, differentiate \( y_2 = \sqrt{1-4x^2} \):
\[ \frac{dy_2}{dx} = -\frac{4x}{\sqrt{1-4x^2}} \]

### Step 5: Find \( \frac{dx}{dy_2} \)
Taking the reciprocal:
\[ \frac{dx}{dy_2} = -\frac{\sqrt{1-4x^2}}{4x} \]

### Step 6: Combine the Results
Now, substituting back into the chain rule:
\[ \frac{dy_1}{dy_2} = \frac{dy_1}{dx} \cdot \frac{dx}{dy_2} = \left(\frac{1}{\sqrt{1 - (4x\sqrt{1-4x^2})^2}} \cdot \frac{4 - 32x^2}{\sqrt{1-4x^2}}\right) \cdot \left(-\frac{\sqrt{1-4x^2}}{4x}\right) \]
Simplifying:
\[ \frac{dy_1}{dy_2} = -\frac{(4 - 32x^2)}{4x \sqrt{1 - (4x\sqrt{1-4x^2})^2}} \]

### Final Result
Thus, the derivative \( \frac{dy_1}{dy_2} \) is:
\[ \frac{dy_1}{dy_2} = -\frac{(4 - 32x^2)}{4x \sqrt{1 - (4x\sqrt{1-4x^2})^2}} \]