Derivative of a trig function?

Question

The trig function is: sin^2(x/6)
I have as an answer:
2cos(x/6) - ((1-x)/(6))

The process I used was chain rule and on the x/6 part of the chain rule I used the quotient rule.

Is this correct, is my procedure correct, if not please tell me where I went wrong, thanks!

Answer 1

2 ⋅ sin (x/6) ⋅ cos (x/6) / 6

Which simplifies with trig identities as:
sin (x/3) / 6
Or algebraically as:
sin (x/6) ⋅ cos (x/6) / 3


No, youre not correct.

I am. ;P

Chain rule is this:
∂/∂x f( g(x) ) = f'( g(x) ) ⋅ g'(x)
That states that the derivative of a compound function is equal to the derivative of the outer one compounded with the undifferentiated inner one... multiplied by the differentiation of the inner one.

If you start with: sin² (x/6)

Differentiation of the outer one:
y = x²
y' = 2x

Differentiation of the middle one
y = sin x
y' = cos x

Differentiation of the most inner one
y = x/6
y' = 1/6

So,
y = sin² (x/6)
y' = ∂/∂x sin² (x/6)
y' = 2 sin (x/6) ⋅ ∂/∂x sin (x/6)
y' = 2 sin (x/6) ⋅ cos (x/6) ⋅ ∂/∂x (x/6)
y' = 2 sin (x/6) ⋅ cos (x/6) ⋅ (1/6)

Answer 2

No, you have to treat the squared as separate from the sin. So you have (f(x))^2, where f(x) = sin(x/6).

Using the chain rule then, you first take the derivative of the ^2 part. That gives you:
2*sin(x/6)

Now, by the chain rule, you need to multiply that by the derivative of the inside part, which is sin(x/6). So that gives you cos(x/6).

So far, you now have:
2*sin(x/6)*cos(x/6)

Now, again by the chain rule, you have to take the derivative of what is inside the cos, that is x/6. You don't need the quotient rule here. Just treat it as (1/6)*x. The derivative of that would be just 1/6.

So in the end, you have:
2*sin(x/6)*cos(x/6)*(1/6)
= (1/3)*sin(x/6)*cos(x/6)

Answer 3

Cogito isn't right either. It's

2 sin(x/6) cos(x/6) (1/6)

The 2 comes from the squaring; sine is to one less power; the cosine is the derivative of sine, and the 1/6 is the derivative of (x/6), the argument of the sine or cosine

You don't need to use quotient rule on x/6. You can, of course: you get

[(6)(1) - (x)(0)]/6^2 = (1/6)

It's "(x)(0)" in the numerator of the derivative because the derivative of 6 with respect to x (or any other variable) is zero; 6 is a constant.

Answer 4

we just finished learning that, but i got something different and it may be right or wrong...i got
2sin(x/6) multiplied cos (x/6) times (1/6) so the final answer is 1/3 sin(x/6) times cos (x/6).
first i rewrote the ques. to 2 (sin(x/6)) times (the derivative of sin(x/6) which is cos (x/6) times (1/6).....that's what i got. i used the chain rule and then the trig rule. you shouldn't use the quotient rule on this one.

Answer 5

You were close, but neither of the answers are right!! It should be "(1/3)sin(x/6)cos(x/6)"....you kind of have to use the chain rule twice here.

Answer 6

I sent a mail to [e-mail]

Here's what he sent back:

sin^2 (x/6)

Apply Chain Rule

1/3 [sin (x/6) cos (x/6)]

Answer 7

hold on,

g = (sin(x/6))^2

u = sin(x/6)

du/dx = (1/6)cos(x/6)

g = u^2

dg/du = 2u = 2sin(x/6)

Chain rule:
dg/dx = (dg/du)(du/dx) = (1/6)cos(x/6) * 2sin(x/6) =

(1/3)cos(x/6)sin(x/6)



the derivative of the inside function is:

Answer 8

you will could use the chain rule. case in point in case you had to discover the spinoff of y = sin(x^2) it may be y' = cos(x^2)*2x = 2xcos(x^2). on your case i'm not sure in case you advise pi/(3x) or (pi/3)x, it might make a distinction contained in the respond. extremely the chain rule says to discover the spinoff of the exterior functionality evaluated on the interior functionality cases the spinoff of the interior functionality.

Answer 9

y = sin ² (x / 6)
y = [ sin (x / 6) ] ²
dy/dx = 2 [ sin (x / 6)] [ (1 / 6) cos(x / 6) ]
dy/dx = (1 / 3) sin(x / 6) cos (x / 6)
dy/dx = (1 / 6) [ 2 sin(x / 6) cos (x / 6)]
dy/dx = (1 / 6) sin (x / 3)

Answer 10

sin^2(x/6)
Let u =x/6
Then du/dx = 1/6
So d/dx (sinu*sinu) = sinucosu + sinucosu =2sinucosu
= sin 2u du/dx =(1/6)* sin (x/3)