find \\( \\frac{d y}{d x} \\) for \\( y=-18 x+7 \\cos x \\).\n\\( \\frac{d}{d x}(-18 x+7 \\cos x)= \\)

Explanation:

Step1: Differentiate term -18x

Using the power rule (\frac{d}{dx}(ax^n)=nax^{n - 1}), for (y=-18x=-18x^1), (\frac{d}{dx}(-18x)=-18\times1\times x^{1 - 1}=-18).

Step2: Differentiate term 7cosx

Using the rule (\frac{d}{dx}(a\cos x)=-a\sin x), for (y = 7\cos x), (\frac{d}{dx}(7\cos x)=7\times(-\sin x)=-7\sin x).

Step3: Apply the sum rule

The sum rule (\frac{d}{dx}(u + v)=\frac{du}{dx}+\frac{dv}{dx}), where (u=-18x) and (v = 7\cos x). So (\frac{d}{dx}(-18x + 7\cos x)=\frac{d}{dx}(-18x)+\frac{d}{dx}(7\cos x)).

Answer:

(-18-7\sin x)