# Thread: Chain Rule

1. ## 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!

2. Originally Posted by Truthbetold
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