Question

Find the derivative of sin^2 4x^3

Answer 1

I'm getting the wrong answer every time.

f'(x) = (2)(sin 4x^3)(cos 4x^3)(12x^2) = (24x^2)(cos 4x^3)(sin 4x^3)

That's the wrong answer, though.

Answer 2

use the chain rule
derivative of sin^2(4x^3) times derivative of 4x^3

=2sin(4x^3)(12x^2)
=sin(8x^3)(12x^2)

Answer 3

so you don't take the derivative of sin 4x^3?

Answer 4

chain rule three times

Answer 5

you take the derivative of sin^2 then copy 4x^3 then take the derivative of 4x^3

Answer 6

Oh my God! Thanks lgbasallote! I'm studying for the test and this stuff is a couple weeks old, so I'm beginning to forget...wow!

Answer 7

\[\frac{d}{dx}\sin^2(4x^3)\] not quite right still

Answer 8

you have to take the derivative of
a) something squared
b) sin of something
c) 4 something cubed

Answer 9

That makes a lot of sense! Geez...I'm in trouble *gulp*...hehe.

Answer 10

@deniatus try this site it will help you review: www.wolframalpha.com

Answer 11

first the square
the derivative of something squared is twice something
the derivative of sine something is cosine something
and the derivative of 4 something cubed is 14 something squared

Answer 12

your answer should be
\[2\times\sin(4x^3)\times \cos(4x^3)\times 12x^2\]

Answer 13

No, it shouldn't satellite73. That's the answer I got as well and it was incorrect.

Answer 14

lgbasallote is correct with his/her answer.

Answer 15

well actually it is correct for the problem you asked

Answer 16

Trust me, I thought so too but it isn't. I was scratching my head as to what I was doing wrong...

Answer 17

it can be cleaned up as
\[24x^2\sin(4x^3)\cos(4x^3)\]

Answer 18

I KNOW...it's incorrect. It's the SAME answer I got each time....

Answer 19

The answer is 12x^2 sin 8x^3

Answer 20

you can take it to the bank, if the question is "find the derivative of
\[\sin^2(4x^3)\] then the answer is
\[24x^2\sin(4x^3)\cos(4x^3)\]

Answer 21

well sure, that is the same thing right?

Answer 22

Yup, that's exactly the problem. However, the answer is different in the back of the book.

Answer 23

to get from one to the other is the double angle formula

Answer 24

they are using the fact that
\[\sin(2x)=2\sin(x)\cos(x)\]

Answer 25

it is the same thing exactly, just used a trig identity to rewrite it and confuse you

Answer 26

in other words
\[2\sin(4x^3)\cos(4x^3)=\sin(8x^3)\] by the double angle formula

Answer 27

Ah so you and I are correct then...

Answer 28

yes of course they are both the same, but trust me you will not get to the answer in the back of hte book directly. you have to first write it out and then use double angle formula