determine the derivative: y=(3-2x^3)^3 can someone show me the steps? I dont know what I'm doing wrong

Question

anonymous · Answer

Make sure you're using the chain rule:
1) bring down the exponent in front and subtract 1 from it$$y'=3(3-2x^3)^2$$
but don't forget to multiply that by the derivative of the inside of the parenthesis. The derivative of :$$3-2x^3$$
is $$-6x^2$$ so multiply the original derivative we got
$$y'=3(3-2x^3)^2$$
by the derivative of the inside of the parenthesis$$-6x^2$$ to get
$$y'=3(3-2x^3)^2(-6x^2)=-18 x^2 (3-2 x^3)^2$$

buggiethebug · Answer

but we're only supposed to use the product rule, sorry that i forgot to mention

anonymous · Answer

$$y=(3-2x^3)^3$$ is the same as $$y=(3-2x^3)(3-2x^3)^2$$
Square out the one that's squared:
$$(3-2x^3)^2=4 x^6-12 x^3+9$$
Giving you $$y=(3-2x^3)(4 x^6-12 x^3+9)$$
Product rule gives you
$$y'=(3-2x^3)(24 x^5-36 x^2)+(4 x^6-12 x^3+9)(-6x^2)$$
a little simplified: $$y'=(-48 x^8+144 x^5-108 x^2)+(-24 x^8+72 x^5-54 x^2)$$
and more: $$y'=-48 x^8+144 x^5-108 x^2-24 x^8+72 x^5-54 x^2)$$
and finally:$$y'=-72 x^8+216 x^5-162 x^2$$
simplifying to:$$y'=-18 x^2 (2 x^3-3)^2$$
Which is the same as the answer I originally posted:$$y'=-18 x^2 (3-2 x^3)^2$$
.

precal · Answer

you will later find out that chain rule is faster then product rule in this specific case

buggiethebug · Answer

thank you!