Question

Question #92a66

Answer 1

`(3sqrttheta(3sintheta-4thetacostheta))/(4sin^3theta)`

Explanation:

`f(theta)=(3sqrt(theta^3))/(2sin^2theta)`

The first step is to recognize that `sqrt(theta^3)=(theta^3)^(1/2)=theta^(3/2)`.

`f(theta)=(3theta^(3/2))/(2sin^2theta)`

Now, use the quotient rule.

`f'(theta)=((3theta^(3/2))'(2sin^2theta)-(3theta^(3/2))(2sin^2theta)')/(2sin^2theta)^2`

Now, finding these derivatives, we see that the first will require the power rule.

`(3theta^(3/2))'=3(3/2theta^(1/2))=9/2theta^(1/2)`

The other will require the chain rule. The issue is that the sine function is squared.

`(2sin^2theta)'=2(2sin^1theta)*(sintheta)'=4sinthetacostheta`

Now plugging these in:

`f'(theta)=(9/2theta^(1/2)(2sin^2theta)-3theta^(3/2)(4sinthetacostheta))/(2sin^2theta)^2`

`f'(theta)=(9theta^(1/2)sin^2theta-12theta^(3/2)sinthetacostheta)/(4sin^4theta)`

`f'(theta)=(3theta^(1/2)sintheta(3sintheta-4thetacostheta))/(4sin^4theta)`

`f'(theta)=(3sqrttheta(3sintheta-4thetacostheta))/(4sin^3theta)`