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: Differentiate \( y_1 \) with respect to \( x \)
Using the chain rule, we differentiate \( y_1 \):
\[ \frac{dy_1}{dx} = \frac{1}{\sqrt{1 - (4x\sqrt{1-4x^2})^2}} \cdot \frac{d}{dx}(4x\sqrt{1-4x^2}) \]

### Step 3: Differentiate \( 4x\sqrt{1-4x^2} \)
Using the product rule:
\[ \frac{d}{dx}(4x\sqrt{1-4x^2}) = 4\sqrt{1-4x^2} + 4x \cdot \frac{1}{2\sqrt{1-4x^2}} \cdot (-8x) = 4\sqrt{1-4x^2} - \frac{16x^2}{\sqrt{1-4x^2}} \]
Combining the terms:
\[ \frac{d}{dx}(4x\sqrt{1-4x^2}) = \frac{4(1-4x^2) - 16x^2}{\sqrt{1-4x^2}} = \frac{4 - 32x^2}{\sqrt{1-4x^2}} \]

### Step 4: Substitute Back into \( \frac{dy_1}{dx} \)
Now substituting back:
\[ \frac{dy_1}{dx} = \frac{1}{\sqrt{1 - (4x\sqrt{1-4x^2})^2}} \cdot \frac{4 - 32x^2}{\sqrt{1-4x^2}} \]

### Step 5: Differentiate \( y_2 \) with respect to \( x \)
Now differentiate \( y_2 \):
\[ \frac{dy_2}{dx} = \frac{d}{dx}(\sqrt{1-4x^2}) = \frac{-8x}{2\sqrt{1-4x^2}} = \frac{-4x}{\sqrt{1-4x^2}} \]

### Step 6: Use the Chain Rule to Find \( \frac{dy_1}{dy_2} \)
Using the chain rule:
\[ \frac{dy_1}{dy_2} = \frac{dy_1/dx}{dy_2/dx} = \frac{\frac{1}{\sqrt{1 - (4x\sqrt{1-4x^2})^2}} \cdot \frac{4 - 32x^2}{\sqrt{1-4x^2}}}{\frac{-4x}{\sqrt{1-4x^2}}} \]
Simplifying:
\[ \frac{dy_1}{dy_2} = -\frac{(4 - 32x^2)}{4x\sqrt{1 - (4x\sqrt{1-4x^2})^2}} \]

### Step 7: Final Expression
Thus, the final expression for the derivative \( \frac{dy_1}{dy_2} \) is:
\[ \frac{dy_1}{dy_2} = -\frac{(4 - 32x^2)}{4x\sqrt{1 - (4x\sqrt{1-4x^2})^2}} \]