Question

f"(x) of f(x) = xcosx

Answer 1

\[\Large f(x)\quad=\quad x \cos x\]\[\Large f'(x)\quad=\quad \color{#008353 }{(x)'}\cos x+ x\color{#008353 }{(\cos x)'}\]Product rule, yes? :)

Answer 2

So what do you get for the first derivative?

Answer 3

f'(x) = (1)(cosx)+ x(-sinx)?

Answer 4

Mmm ok good.\[\Large f'(x)\quad=\quad \cos x-x\sin x\]

Answer 5

Hmm looks like we have to apply the product rule on that second term.
\[\Large f''(x)\quad=\quad \color{#DD4747}{(\cos x)'}-\left[\color{#DD4747}{(x)'}\sin x+x\color{#DD4747}{(\sin x)'}\right]\]

Answer 6

I just did the square brackets to remind you that you need to distribute that negative to each term that you're applying product rule to.

Answer 7

okay so do you get cosx- sinx+xcosx?

Answer 8

\[\Large f''(x)\quad=\quad \color{#DD4747}{(\cos x)'}-\color{#DD4747}{(x)'}\sin x-x\color{#DD4747}{(\sin x)'}\]Derivative of cosx is -sinx,\[\Large f''(x)\quad=\quad \color{royalblue}{-\sin x}-\color{#DD4747}{(x)'}\sin x-x\color{#DD4747}{(\sin x)'}\]Your middle term looks good, derivative of x is 1,\[\Large f''(x)\quad=\quad \color{royalblue}{-\sin x}-\color{royalblue}{(1)}\sin x-x\color{#DD4747}{(\sin x)'}\]Remember to distribute the negative to the last term^
Derivative of sinx gives us cosx,\[\Large f''(x)\quad=\quad \color{royalblue}{-\sin x}-\color{royalblue}{(1)}\sin x-x\color{royalblue}{(\cos x)}\]

Answer 9

okay so is it -xcosx-2sinx?

Answer 10

mmm yah that looks right! :)