The slope of the tangent to the curve `y=int_x^(x^2)cos^(- 1)t^2dt` at `x=1/(root(4)2)` is

Options

A. `((root(4)(8))/(2)-(3)/(4)) pi`
B. `((root(4)(8))/(3)-(1)/(4)) pi`
C. `((root(5)(8))/(4)-(1)/(3)) pi`
D. none of these

Explanation

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 \( x = \frac{1}{\sqrt[4]{2}} \), we will use the Fundamental Theorem of Calculus (Newton-Leibniz formula) to differentiate the integral with respect to \( x \).

### Step-by-Step Solution:

1. **Identify the function**:
We have:
\[ y = \int_x^{x^2} \cos^{-1}(t^2) \, dt \]

2. **Differentiate using the Fundamental Theorem of Calculus**:
According to the theorem, if \( y = \int_{a(x)}^{b(x)} f(t) \, dt \), then:
\[ \frac{dy}{dx} = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x) \]
Here, \( a(x) = x \) and \( b(x) = x^2 \). Therefore, we need to differentiate:
\[ \frac{dy}{dx} = \cos^{-1}((x^2)^2) \cdot \frac{d}{dx}(x^2) - \cos^{-1}(x^2) \cdot \frac{d}{dx}(x) \]
This simplifies to:
\[ \frac{dy}{dx} = \cos^{-1}(x^4) \cdot (2x) - \cos^{-1}(x^2) \cdot (1) \]

3. **Substitute \( x = \frac{1}{\sqrt[4]{2}} \)**:
First, calculate \( x^2 \) and \( x^4 \):
\[ x^2 = \left(\frac{1}{\sqrt[4]{2}}\right)^2 = \frac{1}{\sqrt{2}} \quad \text{and} \quad x^4 = \left(\frac{1}{\sqrt[4]{2}}\right)^4 = \frac{1}{2} \]
Now substitute these values into the derivative:
\[ \frac{dy}{dx} = \cos^{-1}\left(\frac{1}{2}\right) \cdot (2 \cdot \frac{1}{\sqrt[4]{2}}) - \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) \]

4. **Evaluate the inverse cosine values**:
We know:
\[ \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} \quad \text{and} \quad \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4} \]
Substitute these values back into the derivative:
\[ \frac{dy}{dx} = \frac{\pi}{3} \cdot \left(2 \cdot \frac{1}{\sqrt[4]{2}}\right) - \frac{\pi}{4} \]

5. **Simplify the expression**:
The expression becomes:
\[ \frac{dy}{dx} = \frac{2\pi}{3\sqrt[4]{2}} - \frac{\pi}{4} \]
To combine these fractions, find a common denominator, which is \( 12\sqrt[4]{2} \):
\[ \frac{dy}{dx} = \frac{8\pi}{12\sqrt[4]{2}} - \frac{3\pi\sqrt[4]{2}}{12\sqrt[4]{2}} = \frac{8\pi - 3\pi\sqrt[4]{2}}{12\sqrt[4]{2}} \]

6. **Final result**:
The slope of the tangent to the curve at \( x = \frac{1}{\sqrt[4]{2}} \) is:
\[ \frac{dy}{dx} = \frac{\pi(8 - 3\sqrt[4]{2})}{12\sqrt[4]{2}} \]