Question

Find y' and y'' of:
\[y=\cos(\sin3x)\]

I solved for y' and not sure how to go about y''
\[y=\cos(\sin3x) \rightarrow y'=-\cos(3x)\times \sin(\sin3x) \]

Answer 1

y'=-3cos(3x) (sin(sin3x)
Forgot the 3

Answer 2

\[\large\rm y'=(-3\cos3x)\sin(\sin3x)\]Ok your first derivative looks great.
Let's first "set up" our second derivative,\[\large\rm y''=\color{royalblue}{(-3\cos3x)'}\sin(\sin3x)+(-3\cos3x)\color{royalblue}{(\sin(\sin3x))'}\]

Answer 3

Oh I forgot about the product rule. So for the first part, would it be 9sin3x times sin(sin3x)

Answer 4

\[\large\rm y''=\color{orangered}{(9\sin3x)}\sin(\sin3x)+(-3\cos3x)\color{royalblue}{(\sin(\sin3x))'}\]Sounds right so far!

Answer 5

\[9\sin3x \times \sin(\sin3x) + -3\cos3x \times 3\cos(\cos3x)\]

Answer 6

I think the second part might be missing something

Answer 7

\[\large\rm \frac{d}{dx}\sin(\sin3x)=\cos(\sin3x)\cdot\color{royalblue}{(\sin3x)'}\]

Answer 8

\[\large\rm \frac{d}{dx}\sin(\sin3x)=\cos(\sin3x)\cdot\color{orangered}{(3\cos3x)}\]

Answer 9

Ah okay.

Answer 10

So our final answer would be:
\[y"=(9\sin3x)(\sin(\sin3x))+(-3\cos3x)(\cos(\sin3x)(3\cos3x)\]

Answer 11

\[\large\rm y''=\color{orangered}{(9\sin3x)}\sin(\sin3x)+(-3\cos3x)\color{royalblue}{(\sin(\sin3x))'}\]\[\large\rm y''=\color{orangered}{(9\sin3x)}\sin(\sin3x)+(-3\cos3x)\color{orangered}{(\cos(\sin3x)\cdot3\cos3x)}\]Mmm good good.
Gonna clean it up a little bit? :) heh

Answer 12

Oh I guess you can't do very much simplification actually..

Answer 13

Oh okay. That's fine by me if we can't simplify it :D

Answer 14

Thanks for the help!

Answer 15

Bring the 3's together in the second term,
and bring the cos3x's together as well.\[\large\rm y''=9\sin3x\cdot \sin(\sin3x)-9\cos^23x\cdot\cos(\sin3x)\]That looks a little nicer, not a big deal though.

np!