What is the third derivative of y = cos(2θ)

First, note that d/dx cos(f(x)) = -f(x)sin(f(x)) and d/dx sin(f(x)) = -f(x)cos(f(x)), so:

dy/dθ = -2sin(2θ)
d²y/dθ² = -2(2cos(2θ)) = -4cos(2θ)
d³y/dθ³ = -4(-2sin(2θ)) = 8sin(2θ)

I hope this helps!

You must use the chain rule. Find the derivative of u =2θ that is u´(θ) =2 and find the derivative of
cos(u) that is -sin(u) and multiply.

f´(θ)= f´(u) u`(θ) = -sin(2θ)[2]

Then dy/dθ = -2sin(2θ) OK!

we have a formula

'n'th derivative of cos(ax+b) = a^n*cos[ax+b+nπ/2]

here n=3

3rd derivative of cos(2θ)=(2³)*cos[2θ+3π/2] = 8*cos[2θ+3π/2] = 8*sin[2θ]

y = (cot(cos(θ))²

Use the chain rule:

y ' = 20(cot(cos(θ)) (-csc²(cosθ)) (-sin(θ))

y ' = 20(cot(cosθ) (csc²(cosθ)) (sinθ)

the answers B homie.