Question

basic trig derivatives with/without chain rule
this is the only question in this section.
question
find $\frac{d}{dx}(-2\cos x - 4)$

Explanation:

Step1: Apply derivative rules

Use the sum rule \(\frac{d}{dx}(u - v)=\frac{du}{dx}-\frac{dv}{dx}\). Let \(u=-2\cos x\) and \(v = 4\). Then \(\frac{d}{dx}(-2\cos x-4)=\frac{d}{dx}(-2\cos x)-\frac{d}{dx}(4)\).

Step2: Differentiate each term

For \(\frac{d}{dx}(-2\cos x)\), use the constant - multiple rule \(\frac{d}{dx}(k\cdot f(x))=k\frac{df(x)}{dx}\) (\(k = - 2\), \(f(x)=\cos x\)). The derivative of \(\cos x\) is \(-\sin x\), so \(\frac{d}{dx}(-2\cos x)=-2\frac{d}{dx}(\cos x)=-2(-\sin x) = 2\sin x\).
The derivative of a constant \(c\) (here \(c = 4\)) is \(0\), so \(\frac{d}{dx}(4)=0\).

Answer:

\(2\sin x\)