How to find the derivative with a csc in the function

Question

h(t)= 1/csc(4t)

 

I tried this many ways and got different answers, yet none of them were correct. I then tried to raise it to a negative power so I could avoid mistakes using the quotient rule, but i still got it wrong. What am I doing wrong?

Answer 1

Trig Identity: csc(x) = 1/sin(x)

 

h(t) = 1/csc(4t) = sin(4t)

 

h'(t) = 4cos(4t) 

Answer 2

I applied the trigonometric derivative of csc(x).

 

d/dx(csc(x)) = -csc(x)cot(x)

 

 

The derivative of h(t) using the quotient rule

 

h'(t) = [4csc(4t)cot(4t)] / csc²(4t)

 

 

Now just simplify the derivative.

 

h'(t) = [4cos(4t) / sin²(4t)] * sin²(4t)

 

h'(t) = 4cos(4t)