Question

basic trig derivatives with/without chain rule
this is the only question in this section.
question
find \\( \frac{d}{d x}(-\sin (8 x+9)) \\)

Explanation:

Step1: Apply the derivative of a constant multiple

The derivative of \( - \sin(8x + 9)\) is \(-\frac{d}{dx}(\sin(8x + 9))\).

Step2: Apply the chain rule

Let \(u = 8x+9\), then \(\frac{d}{dx}(\sin(u))=\cos(u)\cdot\frac{du}{dx}\).
Since \(\frac{du}{dx}=8\), we have \(\cos(8x + 9)\cdot8\).

Step3: Combine the results

\(-\frac{d}{dx}(\sin(8x + 9))=-8\cos(8x + 9)\)

Answer:

\(-8\cos(8x + 9)\)