Results 1 to 2 of 2

Math Help - Chain Rule

  1. #1
    Member
    Joined
    Oct 2007
    Posts
    178

    Chain Rule

    Find dy/dx

    y= sin^2(3x-2)


    I don't know how to work with sin^2.

    I used the chain rule twice and got 6 cos(3x-2) .
    That, however, is not the correct answer.
    It is 3 sin (6x-4) .

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Truthbetold View Post
    Find dy/dx

    y= sin^2(3x-2)


    I don't know how to work with sin^2.

    I used the chain rule twice and got 6 cos(3x-2) .
    That, however, is not the correct answer.
    It is 3 sin (6x-4) .

    Thanks!
    your answer is way off. the correct answer is not readily apparent, you have to apply a trig identity to get it in the required form.

    the trick is to think of \sin^2 (3x - 2) as (\sin (3x - 2))^2

    so, y = \sin^2 (3x - 2)

    \Rightarrow \frac {dy}{dx} = 2 \sin (3x - 2) \cdot 3 \cos (3x - 2) ..............by the chain rule

    = 6 \sin (3x - 2) \cos (3x - 2)

    = 3 \cdot 2 \sin (3x - 2) \cos (3x - 2)

    now recall the identity, 2 \sin A \cos A = \sin 2A, thus we have:

    \frac {dy}{dx} = 3 \sin 2(3x - 2)

    = 3 \sin (6x - 4)

    as desired
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 9
    Last Post: November 9th 2010, 02:40 AM
  2. Chain Rule Inside of Chain Rule
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 22nd 2009, 09:50 PM
  3. Replies: 5
    Last Post: October 19th 2009, 02:04 PM
  4. Replies: 3
    Last Post: May 25th 2009, 07:15 AM
  5. Replies: 2
    Last Post: December 13th 2007, 06:14 AM

Search Tags


/mathhelpforum @mathhelpforum