Question

If \[y = \frac{\left(\sqrt{x} + 1\right)\left(x^2 - \sqrt{x}\right)}{x\sqrt{x} + x + \sqrt{x}} + \frac{1}{15}(3\cos^2 x - 5)\cos^3 x,\]then $96y'\left(\frac{\pi}{6}\right)$ is equal to:

Answer 1

Correct Answer: 105

Given:

The equation to analyze is: \[ y = \left( \sqrt{x+1} \right) \cdot \frac{x^2 - \sqrt{x}}{\sqrt{x + \sqrt{x}} + x} + \frac{1}{15} \left( 3 \cos^2 x - 5 \right) \cos^3 x \]

Step 1: Equation Simplification:

Simplify the components of the equation: \[ y = \left( \sqrt{x+1} \right) \cdot \sqrt{x(\sqrt{x}-1)} \cdot \frac{x + \sqrt{x+1}}{\sqrt{x(\sqrt{x}+1)}} \] \[ y = (x - 1) + \frac{1}{15} \left( 3 \cos^5 x - 5 \cos^3 x \right) \]

Step 2: Differentiation:

Differentiate the simplified equation with respect to \(x\): \[ y' = 1 - 0 + \frac{1}{15} \left[ 15 \cos^4 x \cdot (- \sin x) - 15 \cos^2 x \cdot (- \sin x) \right] \] The derivative simplifies to: \[ y' = 1 - \sin x \cdot \left( \cos^4 x - \cos^2 x \right) \]

Step 3: Evaluation at \( x = \frac{\pi}{6} \):

Substitute \( x = \frac{\pi}{6} \) into the derivative \( y' \): \[ y'\left( \frac{\pi}{6} \right) = 1 - \sin \left( \frac{\pi}{6} \right) \left[ \cos^4 \left( \frac{\pi}{6} \right) - \cos^2 \left( \frac{\pi}{6} \right) \right] \] Perform the substitution and simplification: \[ y'\left( \frac{\pi}{6} \right) = 1 - \frac{1}{2} \left( \frac{9}{16} - \frac{3}{4} \right) \] \[ y'\left( \frac{\pi}{6} \right) = 1 - \frac{1}{2} \left( \frac{9}{16} - \frac{12}{16} \right) \] \[ y'\left( \frac{\pi}{6} \right) = 1 - \frac{1}{2} \cdot \left( \frac{-3}{16} \right) \] \[ y'\left( \frac{\pi}{6} \right) = 1 + \frac{3}{32} \] \[ y'\left( \frac{\pi}{6} \right) = \frac{35}{32} \]

Step 4: Final Calculation:

Multiply the evaluated derivative by 96: \[ 96 y'\left( \frac{\pi}{6} \right) = 96 \times \frac{35}{32} = 105 \]

Conclusion: The final result is 105.