The slope of the tangent of the curve `y=int_(x)^(x^(2))(cos^(-1)t^(2))dt` at `x=(1)/(sqrt2)` is equal to

Options

A. `cos^(-1)((1)/(4))-(pi)/(3)`
B. `cos^(-1)((1)/(4))+(pi)/(3)`
C. `sqrt2cos^(-1)((1)/(4))-(pi)/(3)`
D. `sqrt2 cos ^(-1)((1)/(4)+(pi)/(3)`

Answer

To find the slope of the tangent of the curve given by the equation

\[ y = \int_{x}^{x^2} \cos^{-1}(t^2) \, dt \]

at the point \( x = \frac{1}{\sqrt{2}} \), we will follow these steps:

### Step 1: Differentiate the integral

Using the Leibniz rule for differentiation under the integral sign, we can differentiate \( y \) with respect to \( x \):

\[ \frac{dy}{dx} = \frac{d}{dx} \left( \int_{x}^{x^2} \cos^{-1}(t^2) \, dt \right) = \cos^{-1}((x^2)^2) \cdot \frac{d}{dx}(x^2) - \cos^{-1}(x^2) \cdot \frac{d}{dx}(x) \]

### Step 2: Compute the derivatives

Calculating the derivatives:

1. \(\frac{d}{dx}(x^2) = 2x\)
2. \(\frac{d}{dx}(x) = 1\)

Thus, we can rewrite the expression for \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} = \cos^{-1}(x^4) \cdot (2x) - \cos^{-1}(x^2) \cdot 1 \]

### Step 3: Substitute \( x = \frac{1}{\sqrt{2}} \)

Now we substitute \( x = \frac{1}{\sqrt{2}} \):

1. Calculate \( x^2 = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \)
2. Calculate \( x^4 = \left(\frac{1}{\sqrt{2}}\right)^4 = \frac{1}{4} \)

Now substitute these values into the derivative:

\[ \frac{dy}{dx} = \cos^{-1}\left(\frac{1}{4}\right) \cdot \left(2 \cdot \frac{1}{\sqrt{2}}\right) - \cos^{-1}\left(\frac{1}{2}\right) \]

### Step 4: Evaluate the inverse cosine values

1. \(\cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}\)
2. \(\cos^{-1}\left(\frac{1}{4}\right)\) remains as it is.

Thus, we have:

\[ \frac{dy}{dx} = \cos^{-1}\left(\frac{1}{4}\right) \cdot \sqrt{2} - \frac{\pi}{3} \]

### Final Result

The slope of the tangent at \( x = \frac{1}{\sqrt{2}} \) is:

\[ \frac{dy}{dx} = \sqrt{2} \cos^{-1}\left(\frac{1}{4}\right) - \frac{\pi}{3} \]