Question

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

\(-18-7\sin x\)