Need help finding the derivative of sin^2 (x/3)

[abel muroi]

I usually don't have a problem finding the derivative of functions, but this particular expression is driving me nuts.

I was given this expression

sin²(x/3)

and I was told to find the derivative.


From what I know, I know that sin²(x/3) is the same as (sin (x/3))²

So I first used the chain rule
f(u) = u² ----> 2u

g(x) = sin (x/3) ------> (1)/(3) cos(x/3) (I used the chain rule again for this expression)


(2 sin (x/3) ) (1/3 cos (x/3)) = (2 sin (x/3) cos (x/3))/(3)


My final answer is (2 sin (x/3) cos(x/3))/(3)

Did i make any mistakes anywhere? if so, can you show me where exactly did i make a mistake?

 

[stapel]

I usually don't have a problem finding the derivative of functions, but this particular expression is driving me nuts.

I was given this expression

sin²(x/3)

and I was told to find the derivative.

One difficulty may stem from the fact that this isn't actually a functional statement; there is no "equals" sign, let alone an equation.

From what I know, I know that sin²(x/3) is the same as (sin (x/3))²

Yes; the squaring is "outside" of the sine (which, in this case, is outside of the (1/3)*(x)).

So I first used the chain rule
f(u) = u² ----> 2u

g(x) = sin (x/3) ------> (1)/(3) cos(x/3) (I used the chain rule again for this expression)

(2 sin (x/3) ) (1/3 cos (x/3)) = (2 sin (x/3) cos (x/3))/(3)

My final answer is (2 sin (x/3) cos(x/3))/(3)

It might be better (in the sense of formatting, rather than anything mathematical) to state this as:

. . . . .$\displaystyle f"(x)\, =\, \dfrac{2}{3}\, \sin\left(\dfrac{x}{3}\right)\, \cos\left(\dfrac{x}{3}\right)$

Otherwise, everything looks fine to me. Thank you for showing your work and reasoning so nicely!