derivatives calculus

Question

Find the derivative of the function.

 

f(θ)= arcsin (√(sin)(9θ))

 

sin(9θ) is in the square root 

Answer 1

You can differentiate this function by using implicit differentiation, provided that you don't know what the derivative of arcsin(x) is, and the chain rule and the fact that ddx(sinx)=cosx.

f(θ)=arcsin(√sin(9θ)) = sin(f)=√sin(9θ)

Differentiate both sides with respect to θ

ddθ(sin(f))=ddθ√sin(9θ)

Find ddθ√sin(9θ) first using the chain rule

ddθ√sin(9θ)=1/2sin(9θ)−12⋅cos(9θ)⋅9

ddθ√sin(9θ)=9/2sin(9θ)−12⋅cos(9θ)

Take this back to your target derivative to get

cos(f)⋅dfdθ=9/2sin(9θ)−12⋅cos(9θ)

Isolate dfdθ on one side

dfdθ=9/2⋅sin(9θ)−12⋅cos(9θ)cos(y)

Use the trigonometric identity sin2x + cos2x =1 to write cosx as a function of sinx:

cos2x=1−sin2x⇒cosx=√1−sin2x

This will get you

dfdθ=9/2⋅sin(9θ)−12⋅cos(9θ)√1−sin2(y)

Finally, use the first equation to get



dfdθ=9/2⋅sin(9θ)−12⋅cos(9θ)dfdθ=92⋅cos(9θ)√sin(9θ)⋅√1−sin(9θ)

This is equivalent to

dfdθ=9/2cos(9θ)√sin(9θ)⋅√1−sin(9θ)