Find the derivative of sin6theta and (1)/((sin^3)theta)?

Answer 1

I do belive that answers are:


derivative: sin(6θ) = 6cos(6θ)

derivative: 1/sin^3(θ) = -3cos(θ)/sin^4(θ)

Answer 2

ok, so for this question, you have to use the chain rule. Basically, this says that du/dt = du/dv * dv/dt

I will start with sin6theta. (I'm going to assume this is sin (6theta) instead of (sin^6) theta for the first part. I will also use 't' instead of 'theta' because its shorter)

So assign the variable u to equal sin (6t), and the variable v = 6t. So then u would equal sin v.

Now you want to find du/dt, which is equal to du/dv * dv/dt.
u = sin v, so du/dv is equal to cos v.
v = 6t, so dv/dt = 6.

So then du/dt = 6*cosv. Now you have to put the v=6t back in the u equation and you will get:

du/dt = 6*cos (6t).

[If the question was (sin^6) t, then you would set v = sin t, and then u would be v^6 since sin^6 t = (sin t)^6.

So then du/dt = du/dv * dv/dt, and du/dv = 6v^5 and dv/dt = cos t, so then du/dt = 6v^5 * cos t = 6 sin^5 t * cos t.]

For the second part, you can change 1/(sin^3 t) to sin^(-3) t. and to almost the same thing. Set v = sin t and u = sin^(-3) t = v^(-3).

du/dv = -3*v^(-4); dv/dt = cos t,

du/dt = -3*cos t * v^-4 = -3*cos t * (sin t)^(-4) = -3 cos t / sin^4 t

Hope that helped.

Answer 3

1) f(t) = sin(6t). By the chain rule,
f'(t) = 6cos(6t)

2) f(t) = 1/[sin^3(t)]

By identities, since sin(t) = 1/csc(t), csc(t) = 1/sin(t).

f(t) = csc^3(t)

f'(t) = 3[csc^2(t)] [-csc(t)cot(t)]

Which simplifies as

f'(t) = -3 csc^3(t)cot(t)

Answer 4

Question 1
y = sin 6Ø
let u = 6Ø so y = sin u
du/dØ = 6
dy/du = cos u = cos 6Ø
dy / dØ = (du/dØ).(dy/du) = 6.cos.6Ø

Question 2
y = sin^(-3)Ø = (sinØ)^(- 3)
dy/dØ = (-3).sin^(-4).Ø.cos Ø
dy/dØ = (-3).cosØ / sin^(4)Ø