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Thread: Chain Rule

  1. #1
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    Chain Rule

    Find dy/dx

    $\displaystyle y= sin^2(3x-2)$


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

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

    Thanks!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Truthbetold View Post
    Find dy/dx

    $\displaystyle y= sin^2(3x-2)$


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

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

    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 $\displaystyle \sin^2 (3x - 2)$ as $\displaystyle (\sin (3x - 2))^2$

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

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

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

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

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

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

    $\displaystyle = 3 \sin (6x - 4)$

    as desired
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