Question

So I'm looking for some help on Calculus A, chapter 2 on differentiation. I know how to do the gist of it but i have difficulty with the chain rule.
y=x²sine³(2x)

Answer 1

Let's do this step by step?

Answer 2

I know that in the chain rule you use \[f \prime( g(x))\times g \prime(x)\]

Answer 3

i know the following... well i figured it out:
f(x)=x²(u)
g(x)=sin³(x)
f’(x)= 3u²
g’(x)=3cos²x

Answer 4

That is correct
But you see here we have a chain rule and a product rule

Answer 5

well you do all the chain rule of sin³(2x) and then you just do product rule with x² and whatever you got with the product rule right??

Answer 6

Yes exactly, do that now
and ill check if you're right

Answer 7

okay so i have so far that the chain rule thing is
\[3(\sin ^{3}(x))^{2}(3cosx)\]
which simplifies to
\[(3\sin ^{5} x)(3\cos²x)\]
and then
\[9\sin ^{5} xcos²x\]

Answer 8

and now is where i stop because i am lost and don't know how to do the rest.... in fact I'm not even sure i did that part right

Answer 9

That is incorrect
The chain rule is as follows
\[\frac{ d }{dx } (\sin(2x))^3\]

Answer 10

And then....

Answer 11

hommie where did you learn your chain rule?!?!

Answer 12

thats implicit differentiation not the chain rule!!

Answer 13

\[3(\sin(2x))^3 \times 2(\sin(2x)) \times \cos(2x)\]

Answer 14

never mind i get it

Answer 15

actually, no this is the chain rule
Implicit is taking the derivative of y too.

Answer 16

why is it that and not f '(g(x))(g '(x)

Answer 17

because you just rearranged the original equation and got a whole bunch of stuff out of it

Answer 18

Well no you have repeated chain rule for this case.
I didnt rearrange the equation
I just took the derivative for the chain rule term
You're still not done
You gotta do the product rule

Answer 19

like i get how you got all the numbers but how do you know when to simplify it the way you did and when to do it the way i do?

Answer 20

i know.... let me finish the problem on paper and get back to you in a bit.

Answer 21

Well you look at whether you took the derivative of every function within the chain of functions
you have to know your f(x) and g(x) when it comes to f(x) and g(x)
Btw I did it using the actual way which is the "way you do"
Like i said, you have repeated chain rule.

Answer 22

are you absolutely sure your solution is correct? or am i just going crazy and doing it completely wrong?

Answer 23

Proof^

Answer 24

what the heck like literally everything i do is not that answer.....

Answer 25

ohh goodness i thought i was going crazy.....
\[2(3\cos ^{2}(x)\sin ^{3}(2x)\]

Answer 26

Not it

Answer 27

i know this

Answer 28

I would explain but I'm pretty tired

Answer 29

well i don't get it but its fine ill figure it out thanks for the help.