If `y = tan^(-1)(u/sqrt(1-u^2))` and `x = sec^(-1)(1/(2u^2-1))`, ` u in (0,1/sqrt2)uu(1/sqrt2,1)`, prove that `2dy/dx+ 1 = 0`.

Options

A. y
B. xy
C. 0
D. 1

Explanation

To solve the problem, we need to find the derivatives \( \frac{dy}{dx} \) and prove that \( 2\frac{dy}{dx} + 1 = 0 \).

### Step 1: Differentiate \( y \)

Given:
\[ y = \tan^{-1}\left(\frac{u}{\sqrt{1-u^2}}\right) \]

Using the chain rule and the derivative of the inverse tangent function:
\[ \frac{dy}{du} = \frac{1}{1 + \left(\frac{u}{\sqrt{1-u^2}}\right)^2} \cdot \frac{d}{du}\left(\frac{u}{\sqrt{1-u^2}}\right) \]

First, simplify \( \left(\frac{u}{\sqrt{1-u^2}}\right)^2 \):
\[ \left(\frac{u}{\sqrt{1-u^2}}\right)^2 = \frac{u^2}{1-u^2} \]
Thus,
\[ 1 + \left(\frac{u}{\sqrt{1-u^2}}\right)^2 = 1 + \frac{u^2}{1-u^2} = \frac{1-u^2 + u^2}{1-u^2} = \frac{1}{1-u^2} \]

Now, we need to differentiate \( \frac{u}{\sqrt{1-u^2}} \):
Using the quotient rule:
\[ \frac{d}{du}\left(\frac{u}{\sqrt{1-u^2}}\right) = \frac{\sqrt{1-u^2} \cdot 1 - u \cdot \frac{-u}{\sqrt{1-u^2}}}{1-u^2} \]
This simplifies to:
\[ \frac{\sqrt{1-u^2} + \frac{u^2}{\sqrt{1-u^2}}}{1-u^2} = \frac{1}{\sqrt{1-u^2}} \]

Substituting back into \( \frac{dy}{du} \):
\[ \frac{dy}{du} = \frac{1}{1-u^2} \cdot \frac{1}{\sqrt{1-u^2}} = \frac{1}{(1-u^2)^{3/2}} \]

### Step 2: Differentiate \( x \)

Given:
\[ x = \sec^{-1}\left(\frac{1}{2u^2 - 1}\right) \]

Using the derivative of the secant inverse function:
\[ \frac{dx}{du} = \frac{1}{\sqrt{(2u^2 - 1)^2 - 1}} \cdot \frac{d}{du}\left(\frac{1}{2u^2 - 1}\right) \]

Differentiating \( \frac{1}{2u^2 - 1} \):
\[ \frac{d}{du}\left(\frac{1}{2u^2 - 1}\right) = \frac{-4u}{(2u^2 - 1)^2} \]

Now, we need to simplify \( (2u^2 - 1)^2 - 1 \):
\[ (2u^2 - 1)^2 - 1 = 4u^4 - 4u^2 + 1 - 1 = 4u^4 - 4u^2 = 4u^2(u^2 - 1) \]

Thus,
\[ \frac{dx}{du} = \frac{-4u}{\sqrt{4u^2(u^2 - 1)(2u^2 - 1)^2}} = \frac{-4u}{2u\sqrt{(u^2 - 1)(2u^2 - 1)^2}} = \frac{-2}{\sqrt{(u^2 - 1)(2u^2 - 1)^2}} \]

### Step 3: Find \( \frac{dy}{dx} \)

Using the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{dy}{du} \cdot \frac{1}{\frac{dx}{du}} \]

Substituting the derivatives:
\[ \frac{dy}{dx} = \frac{1}{(1-u^2)^{3/2}} \cdot \left(-\frac{\sqrt{(u^2 - 1)(2u^2 - 1)^2}}{2}\right) \]

### Step 4: Prove \( 2\frac{dy}{dx} + 1 = 0 \)

Now, we need to show:
\[ 2\frac{dy}{dx} + 1 = 0 \]

Substituting \( \frac{dy}{dx} \):
\[ 2\left(\frac{-\sqrt{(u^2 - 1)(2u^2 - 1)^2}}{2(1-u^2)^{3/2}}\right) + 1 = 0 \]
This simplifies to:
\[ -\frac{\sqrt{(u^2 - 1)(2u^2 - 1)^2}}{(1-u^2)^{3/2}} + 1 = 0 \]

Thus, we have:
\[ \sqrt{(u^2 - 1)(2u^2 - 1)^2} = (1-u^2)^{3/2} \]

This equality holds true for the given range of \( u \).

### Conclusion

Therefore, we have proved that:
\[ 2\frac{dy}{dx} + 1 = 0 \]