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

Step 1. Simplify the Expression for y:

The problem starts with a complex expression for y. Through algebraic manipulation (not explicitly shown in the image, but implied by the result), this simplifies to a much cleaner form:
\(y = (x - 1) + \frac{1}{15} [3 \cos^5 x - 5 \cos^3 x]\)

Step 2. Differentiate with Respect to x:

We need to find the derivative of y with respect to x, denoted as y'. We differentiate term by term, using the chain rule for the trigonometric terms:
\(y' = \frac{d}{dx} \left[ (x - 1) + \frac{1}{15} (3 \cos^5 x - 5 \cos^3 x) \right]\)

This gives:
\(y' = 1 + \frac{1}{15} [15 \cos^4 x (-\sin x) - 15 \cos^2 x (-\sin x)]\)

Simplifying:
\(y' = 1 - \sin x [\cos^4 x - \cos^2 x]\)

Step 3. Evaluate y'(π/6):

We substitute x = π/6 into the expression for y':
\(y'(\pi/6) = 1 - \sin(\pi/6) \left[ \cos^4(\pi/6) - \cos^2(\pi/6) \right]\)

We know that $\sin(\pi/6) = 1/2$ and $\cos(\pi/6) = \frac{\sqrt{3}}{2}$. Substituting these values:
\(y'(\pi/6) = 1 - \frac{1}{2} \left[ \left(\frac{\sqrt{3}}{2}\right)^4 - \left(\frac{\sqrt{3}}{2}\right)^2 \right] = 1 - \frac{1}{2} \left[ \frac{9}{16} - \frac{3}{4} \right]\)

Simplifying the fraction:
\(y'(\pi/6) = 1 - \frac{1}{2} \left[ \frac{9 - 12}{16} \right] = 1 + \frac{3}{32} = \frac{35}{32}\)

Step 4. Final Calculation:

Finally, we compute 96 \(\times\) y'(π/6):
\(96 \cdot y'(\pi/6) = 96 \times \frac{35}{32} = 3 \times 35 = 105\)

Therefore, the final answer is:

$\boxed{105}$