Step 1: Use the chain rule for differentiation. Let \( u = \sin x^2 \), so that \( \cos(\sin x^2) = \cos(u) \).
Step 2: Differentiate \( \cos(u) \) with respect to \( u \): \[ \frac{d}{du} \cos(u) = -\sin(u) \] Thus, the derivative of \( \cos(\sin x^2) \) is \( -\sin(\sin x^2) \).
Step 3: Now, differentiate \( \sin x^2 \) with respect to \( x \) using the chain rule: \[ \frac{d}{dx} \sin x^2 = 2x \cos x^2 \]
Step 4: Multiply the results from the previous steps: \[ \frac{d}{dx} \cos(\sin x^2) = -2x \sin(2x^2) \cos(\sin x^2) \]
Find the interval in which $f(x) = x + \frac{1}{x}$ is always increasing, $x \neq 0$.
Find the values of \( x, y, z \) if the matrix \( A \) satisfies the equation \( A^T A = I \), where
\[ A = \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix} \]