Question

I have a question about product rule and chain rule. For example the derivative of sin(2x) should be cos(2x) * 2, so what if you had xcos(2x)? Would it be x * cos(2x) + (1 * sin(2x)) * 2 or would it be (x*cos(2)?)

Answer 1

If you had xcos(2x) you need the product rule: So:
\[\frac{d}{dx}f(x)*g(x)=f'(x)g(x)+f(x)g'(x)\]
Let: f(x)=x, g(x)=cos(2x)
then:
f'(x)=1, g'(x)=-2sin(2x)
So:
\[\frac{d}{dx}x \cos(2x)=(1)(\cos(2x))+(x)(-2\sin(2x))=\cos(2x)-2x \sin(2x)\]

Answer 2

The -2 on g'(x) comes from the chain rule, so:
\[\frac{d}{dx}f(g(x))=g'(x)f'(g(x))\]
Let: f(x)=cos(x)
let g(x)=2x
then: f(g(x))=cos(2x) (right?)
So:
\[\frac{d}{dx}\cos(2x)=(2x)'(\cos(x))'|_{x=2x} \rightarrow 2(-\sin(x))|_{x=2x} \rightarrow -2\sin(2x)\]

Answer 3

Ah I see, I combine them.... I was confused at what the professor did on the board, it seemed like they didn't add the chain rule in so I got confused. Thanks :)

Answer 4

No problem :P