If `y=sec^(-1)(1/(4x^(3)-3x)),"find "dy/dx`.

Answer

To find \(\frac{dy}{dx}\) for the function \(y = \sec^{-1}\left(\frac{1}{4x^3 - 3x}\right)\), we can follow these steps:

### Step 1: Substitute \(x\) with \(\cos \theta\)
Let us set \(x = \cos \theta\). Then, we can express \(y\) in terms of \(\theta\):
\[ y = \sec^{-1}\left(\frac{1}{4(\cos \theta)^3 - 3(\cos \theta)}\right) \]

### Step 2: Simplify the expression inside the secant inverse
Using the identity for the cosine of triple angles, we know:
\[ 4\cos^3 \theta - 3\cos \theta = \cos(3\theta) \]
Thus, we can rewrite \(y\):
\[ y = \sec^{-1}\left(\frac{1}{\cos(3\theta)}\right) \]

### Step 3: Use the secant identity
Since \(\sec^{-1}(x) = \theta\) when \(x = \sec(\theta)\), we have:
\[ y = 3\theta \]

### Step 4: Express \(\theta\) in terms of \(x\)
From our substitution \(x = \cos \theta\), we can express \(\theta\) as:
\[ \theta = \cos^{-1}(x) \]
Thus, substituting back, we get:
\[ y = 3\cos^{-1}(x) \]

### Step 5: Differentiate \(y\) with respect to \(x\)
Now we differentiate \(y\) with respect to \(x\):
\[ \frac{dy}{dx} = 3 \cdot \frac{d}{dx}(\cos^{-1}(x)) \]
Using the derivative of \(\cos^{-1}(x)\):
\[ \frac{d}{dx}(\cos^{-1}(x)) = -\frac{1}{\sqrt{1 - x^2}} \]
Thus:
\[ \frac{dy}{dx} = 3 \cdot \left(-\frac{1}{\sqrt{1 - x^2}}\right) = -\frac{3}{\sqrt{1 - x^2}} \]

### Final Answer
The derivative \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = -\frac{3}{\sqrt{1 - x^2}} \]

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