
Derivative of Sin
I have a problem and to check that I have it correct as I do not have answers.
y=8x sin(3x^42)^3  the product rule kicks in here
y'=8 sin(3x^42)^3 + 8x cos(3x^42)^3 (3)(3x^42)^2 (12x^3)
Can someone check to see if I understand this please?
Thanks
Joanne

sin(3x^42)^3 can mean
$\displaystyle \sin^3(3x^4  2)\ \rm{i.e.}\m\ [\sin(3x^4  2)]^3$
and if that's what you mean then your working is good except for a missing 'sin' between the (3) and the subsequent (3x^42). Edit: sorry, no you need to remove the power 3 from the cos, as well. In that case I'll do a pic, as well.
If sin(3x^42)^3 means
$\displaystyle \sin[(3x^42)^3]$ then that's another matter. [Oh... you did mean that, and it's all good.]

$\displaystyle u = 8x \rightarrow u' = 8$
$\displaystyle v = \sin(3x^42)^3 \rightarrow v' = \cos(3x^42)^3 \cdot 36x^3(3x^42)^2$
$\displaystyle y' = vu' + uv' = 8 \cdot (3x^42)^3 + 8x \cdot \cos(3x^42)^3 \cdot 36x^3(3x^42)^2$
$\displaystyle y' = 8(3x^42)^2 [(3x^42) + 36x^4(3x^42) \cos(3x^42)^3$
I do get the same answer as you although I imagine you'd be expected to simplify your equation

Thank you to you both for the reply. It helped.