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:
- \( y_1 = \sin^{-1}(4x\sqrt{1-4x^2}) \)
- \( y_2 = \sqrt{1-4x^2} \)

We need to find \( \frac{dy_1}{dy_2} \).

### Step 2: Differentiate \( y_1 \) with respect to \( x \)
To differentiate \( y_1 \), we will use the chain rule. The derivative of \( \sin^{-1}(u) \) is given by \( \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx} \), where \( u = 4x\sqrt{1-4x^2} \).

First, we find \( u \):
\[ u = 4x\sqrt{1-4x^2} \]

Now, we differentiate \( u \) with respect to \( x \):
\[ \frac{du}{dx} = 4\sqrt{1-4x^2} + 4x \cdot \frac{d}{dx}(\sqrt{1-4x^2}) \]
Using the chain rule for \( \sqrt{1-4x^2} \):
\[ \frac{d}{dx}(\sqrt{1-4x^2}) = \frac{1}{2\sqrt{1-4x^2}} \cdot (-8x) = -\frac{4x}{\sqrt{1-4x^2}} \]
Thus,
\[ \frac{du}{dx} = 4\sqrt{1-4x^2} - \frac{16x^2}{\sqrt{1-4x^2}} = \frac{4(1-4x^2) - 16x^2}{\sqrt{1-4x^2}} = \frac{4 - 16x^2}{\sqrt{1-4x^2}} = \frac{4(1-4x^2)}{\sqrt{1-4x^2}} \]

Now, substituting back into the derivative of \( y_1 \):
\[ \frac{dy_1}{dx} = \frac{1}{\sqrt{1 - (4x\sqrt{1-4x^2})^2}} \cdot \frac{du}{dx} = \frac{1}{\sqrt{1 - 16x^2(1-4x^2)}} \cdot \frac{4(1-4x^2)}{\sqrt{1-4x^2}} \]

### Step 3: Simplify \( \frac{dy_1}{dx} \)
Now we simplify \( 1 - 16x^2(1-4x^2) \):
\[ 1 - 16x^2 + 64x^4 = (1 - 16x^2 + 64x^4) \]
Thus,
\[ \frac{dy_1}{dx} = \frac{4(1-4x^2)}{\sqrt{(1 - 16x^2 + 64x^4)(1-4x^2)}} \]

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

### Step 5: Find \( \frac{dy_1}{dy_2} \)
Now, we can find \( \frac{dy_1}{dy_2} \) using the chain rule:
\[ \frac{dy_1}{dy_2} = \frac{dy_1/dx}{dy_2/dx} = \frac{\frac{4(1-4x^2)}{\sqrt{(1 - 16x^2 + 64x^4)(1-4x^2)}}}{-\frac{4x}{\sqrt{1-4x^2}}} \]
This simplifies to:
\[ \frac{dy_1}{dy_2} = -\frac{(1-4x^2)\sqrt{1-4x^2}}{x\sqrt{1 - 16x^2 + 64x^4}} \]

### Final Result
Thus, the derivative of \( \sin^{-1}(4x\sqrt{1-4x^2}) \) with respect to \( \sqrt{1-4x^2} \) is:
\[ \frac{dy_1}{dy_2} = -\frac{(1-4x^2)\sqrt{1-4x^2}}{x\sqrt{1 - 16x^2 + 64x^4}} \]