# Derivative of Sin

• February 19th 2011, 01:50 PM
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^4-2)^3 - the product rule kicks in here
y'=8 sin(3x^4-2)^3 + 8x cos(3x^4-2)^3 (3)(3x^4-2)^2 (12x^3)

Can someone check to see if I understand this please?

Thanks
Joanne
• February 19th 2011, 01:58 PM
tom@ballooncalculus
sin(3x^4-2)^3 can mean

$\sin^3(3x^4 - 2)\ \rm{i.e.}\m\ [\sin(3x^4 - 2)]^3$

and if that's what you mean then your working is good except for a missing 'sin' between the (3) and the subsequent (3x^4-2). 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^4-2)^3 means

$\sin[(3x^4-2)^3]$ then that's another matter. [Oh... you did mean that, and it's all good.]
• February 19th 2011, 02:00 PM
e^(i*pi)
$u = 8x \rightarrow u' = 8$

$v = \sin(3x^4-2)^3 \rightarrow v' = \cos(3x^4-2)^3 \cdot 36x^3(3x^4-2)^2$

$y' = vu' + uv' = 8 \cdot (3x^4-2)^3 + 8x \cdot \cos(3x^4-2)^3 \cdot 36x^3(3x^4-2)^2$

$y' = 8(3x^4-2)^2 [(3x^4-2) + 36x^4(3x^4-2) \cos(3x^4-2)^3$

I do get the same answer as you although I imagine you'd be expected to simplify your equation
• February 19th 2011, 02:37 PM