Question

@zepdrix Find y" if y=8x sin x

Answer 1

do you know product rule?

Answer 2

Were you able to find the first derivative ok? :o

Answer 3

I get confused when trigs are involved :( is it like 8 sin x + 8x cos x?

Answer 4

\[\large y'=\color{royalblue}{(8x)'}\sin x+8x\color{royalblue}{(\sin x)'}\]
Yah that looks right :)

Answer 5

Awesome!

Answer 6

\[\large y'=8 \sin x+8x \cos x\]Taking the derivative again, this first term shouldn't give you too much trouble, right?
For the other term, you'll repeat this product rule process :O

Answer 7

\[\int\limits (\int\limits (16 \text{Cos}[x]-8 x \text{Sin}[x]) \, dx) \, dx=8 x \text{Sin}[x] \]

Answer 8

\[\large y''=\color{royalblue}{(8\sin x)'}+\color{royalblue}{(8x)'}\cos x+8x\color{royalblue}{(\cos x)'}\]

Answer 9

okay!

Answer 10

Okay so I got 8x sin x + 16 cos x! It should be -8x sin x so where does the negative come from?

Answer 11

The derivative of that last term, cosine x should produce -sine x

Answer 12

Got it! Okay just one last one I promise and you will be free of me!

Answer 13

LOL no more XDDD

Answer 14

LOL okay!