`f'(sin^(2)x)lt f'(cos^(2)x)` for `x in`

Options

A. `(-(pi)/(4),(pi)/(4))`
B. `(-(pi)/(2),-(pi)/(4))uu(pi)/(4),(pi)/(2))`
C. `(-(pi)/(4),0)uu(pi)/(4),(pi)/(2))`
D. `(-(pi)/(2),(pi)/(2))`

Explanation

To solve the inequality \( f'(\sin^2 x) < f'(\cos^2 x) \) for \( x \), we can follow these steps:

### Step 1: Understand the Given Inequality
We start with the inequality:
\[ f'(\sin^2 x) < f'(\cos^2 x) \]
This means that the derivative of the function \( f \) evaluated at \( \sin^2 x \) is less than the derivative evaluated at \( \cos^2 x \).

### Step 2: Relate the Inequality to Trigonometric Functions
From the inequality, we can derive:
\[ \sin^2 x < \cos^2 x \]
This is because if \( f' \) is increasing, then the input must be such that the left side is less than the right side.

### Step 3: Use the Pythagorean Identity
Using the identity \( \sin^2 x + \cos^2 x = 1 \), we can rewrite the inequality:
\[ \sin^2 x < \cos^2 x \implies \sin^2 x < 1 - \sin^2 x \]
This simplifies to:
\[ 2\sin^2 x < 1 \implies \sin^2 x < \frac{1}{2} \]

### Step 4: Solve for \( \sin x \)
Taking the square root of both sides, we get:
\[ -\frac{1}{\sqrt{2}} < \sin x < \frac{1}{\sqrt{2}} \]

### Step 5: Find the Corresponding Angles
The values of \( x \) for which \( \sin x < \frac{1}{\sqrt{2}} \) correspond to the angles:
\[ -\frac{\pi}{4} < x < \frac{\pi}{4} \]
This is because \( \sin x = \frac{1}{\sqrt{2}} \) at \( x = \frac{\pi}{4} \) and \( x = -\frac{\pi}{4} \).

### Step 6: Conclusion
Thus, the interval for \( x \) where the original inequality holds true is:
\[ x \in \left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \]