Derivatice of ` sec^(-1)"" { 1 //(2x^2 -1 } w.r.t sqrt""(1+ 3x )` at ` x=- 1/3` is

Options

A. 0
B. `1//2`
C. `1//3`
D. None

Answer

To find the derivative of \( y = \sec^{-1} \left( \frac{1}{2x^2 - 1} \right) \) with respect to \( z = \sqrt{1 + 3x} \) at \( x = -\frac{1}{3} \), we will follow these steps:

### Step 1: Define the Functions
Let:
- \( y = \sec^{-1} \left( \frac{1}{2x^2 - 1} \right) \)
- \( z = \sqrt{1 + 3x} \)

### Step 2: Differentiate \( y \) with respect to \( x \)
Using the chain rule, the derivative of \( y \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \frac{d}{dx} \left( \sec^{-1}(u) \right) \cdot \frac{du}{dx} \]
where \( u = \frac{1}{2x^2 - 1} \).

The derivative of \( \sec^{-1}(u) \) is:
\[ \frac{d}{du} \left( \sec^{-1}(u) \right) = \frac{1}{|u| \sqrt{u^2 - 1}} \]

Now, we need to find \( \frac{du}{dx} \):
\[ u = \frac{1}{2x^2 - 1} \implies \frac{du}{dx} = -\frac{2x}{(2x^2 - 1)^2} \]

Now substituting \( u \) and \( \frac{du}{dx} \) into the derivative of \( y \):
\[ \frac{dy}{dx} = \frac{1}{\left| \frac{1}{2x^2 - 1} \right| \sqrt{\left( \frac{1}{2x^2 - 1} \right)^2 - 1}} \cdot \left( -\frac{2x}{(2x^2 - 1)^2} \right) \]

### Step 3: Simplify \( \frac{dy}{dx} \)
Now we simplify:
\[ \frac{dy}{dx} = -\frac{2x}{(2x^2 - 1)^2} \cdot \frac{(2x^2 - 1)}{\left| 2x^2 - 1 \right| \sqrt{\frac{1}{(2x^2 - 1)^2} - 1}} \]
This can be further simplified, but let's keep it in this form for now.

### Step 4: Differentiate \( z \) with respect to \( x \)
Now we differentiate \( z \):
\[ z = \sqrt{1 + 3x} \implies \frac{dz}{dx} = \frac{3}{2\sqrt{1 + 3x}} \]

### Step 5: Use the Chain Rule to Find \( \frac{dy}{dz} \)
Using the chain rule:
\[ \frac{dy}{dz} = \frac{dy/dx}{dz/dx} = \frac{\frac{dy}{dx}}{\frac{dz}{dx}} \]

### Step 6: Evaluate at \( x = -\frac{1}{3} \)
Now we substitute \( x = -\frac{1}{3} \) into our expressions for \( \frac{dy}{dx} \) and \( \frac{dz}{dx} \).

1. Calculate \( z \):
\[ z = \sqrt{1 + 3(-\frac{1}{3})} = \sqrt{1 - 1} = 0 \]

2. Calculate \( \frac{dy}{dx} \) at \( x = -\frac{1}{3} \):
\[ u = \frac{1}{2(-\frac{1}{3})^2 - 1} = \frac{1}{\frac{2}{9} - 1} = \frac{1}{-\frac{7}{9}} = -\frac{9}{7} \]
Now substituting \( u \) into \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = -\frac{2(-\frac{1}{3})}{(2(-\frac{1}{3})^2 - 1)^2} \cdot \frac{1}{\left| -\frac{9}{7} \right| \sqrt{\left(-\frac{9}{7}\right)^2 - 1}} \]

3. Calculate \( \frac{dz}{dx} \):
\[ \frac{dz}{dx} = \frac{3}{2\sqrt{1 + 3(-\frac{1}{3})}} = \frac{3}{2\sqrt{0}} \text{ (undefined)} \]

### Step 7: Conclusion
Since \( z \) becomes \( 0 \), \( \frac{dz}{dx} \) is undefined at \( x = -\frac{1}{3} \). Therefore, \( \frac{dy}{dz} \) also becomes undefined.

### Final Answer
The derivative \( \frac{dy}{dz} \) at \( x = -\frac{1}{3} \) is undefined.

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