Question

find \\( \frac { d y } { d x } \\) for \\( y = - 7 x + 3 \cos x \\).
\\( \frac { d } { d x } ( - 7 x + 3 \cos x ) = \square \\)

Explanation:

Step1: Differentiate term -7x

Use the power rule \(\frac{d}{dx}(ax^n)=anx^{n - 1}\). For \(y=-7x=-7x^1\), \(\frac{d}{dx}(-7x)=-7\times1\times x^{1 - 1}=-7\).

Step2: Differentiate term 3cosx

Use the rule \(\frac{d}{dx}(a\cos x)=-a\sin x\). For \(y = 3\cos x\), \(\frac{d}{dx}(3\cos x)=3\times(-\sin x)=-3\sin x\).

Step3: Apply the sum rule of differentiation

If \(y = u + v\), then \(\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}\). Here \(u=-7x\) and \(v = 3\cos x\). So \(\frac{d}{dx}(-7x + 3\cos x)=\frac{d}{dx}(-7x)+\frac{d}{dx}(3\cos x)\).

Answer:

\(-7-3\sin x\)