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 to the curve defined by the integral \( y = \int_{x}^{x^2} \cos^{-1}(t^2) \, dt \) at \( x = \frac{1}{\sqrt{2}} \), we will follow these steps:

### Step 1: Differentiate \( y \) with respect to \( x \)

Using the Fundamental Theorem of Calculus and the Leibniz rule for differentiation under the integral sign, we have:

\[ \frac{dy}{dx} = \frac{d}{dx} \left( \int_{x}^{x^2} \cos^{-1}(t^2) \, dt \right) = \cos^{-1}(x^4) \cdot \frac{d}{dx}(x^2) - \cos^{-1}(x^2) \cdot \frac{d}{dx}(x) \]

### Step 2: Compute the derivatives

Calculating the derivatives:

- The derivative of \( x^2 \) is \( 2x \).
- The derivative of \( x \) is \( 1 \).

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 will substitute \( x = \frac{1}{\sqrt{2}} \) into the derivative:

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

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

Now substituting these values into the derivative:

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

### Step 4: Evaluate the cosine inverse values

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

### Step 5: Final expression for the slope

Putting it all together:

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

Simplifying \( \frac{2}{\sqrt{2}} = \sqrt{2} \):

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

### Final Result

Thus, the slope of the tangent to the curve at \( x = \frac{1}{\sqrt{2}} \) is:

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