Answer is: option1

\( g(x) - 6 = g''''(x)\)

Solution:

Using the identity \( \cos(-x) = \cos x \), we rewrite:

\[ g(x) = \cos x - \sin x + 6 \]

First derivative:

\[ g'(x) = \frac{d}{dx} (\cos x - \sin x + 6) \] \[ = -\sin x - \cos x \]

Second derivative:

\[ g''(x) = \frac{d}{dx} (-\sin x - \cos x) \] \[ = -\cos x + \sin x \]

Third derivative:

\[ g'''(x) = \frac{d}{dx} (-\cos x + \sin x) \] \[ = \sin x + \cos x \]

Fourth derivative:

\[ g''''(x) = \frac{d}{dx} (\sin x + \cos x) \] \[ = \cos x - \sin x \]

Observing that:

\[ g''''(x) = g(x) - 6 \]

Thus, we conclude:

\[ g(x) - 6 = g''''(x) \]