derivative of √cos(t)sect(t)?

Question

the final answer i got was 0! i think i did something wrong..help would be much appreciated

Answer 1

You've gotten different answers from people because it isn't clear whether the radical covers the sec(t) term or not!

Answer 2

The question is analogous to (cos(t)sec(t))^1/2
first use the power rule
1/2(cos(t)sec(t))^-1/2 times the derivative of cossec which we can find using the product rule:
-sin(t)sec(t)+ sec(t)sec(t)tan(t)
so you really have 1/2(cos(t)sec(t))^-1/2 times (-sin(t)sec(t)+sec(t)^2tan(t) which can be simplified to

-sin(t)sec(t)+sec(t)^2tan(t) all divided by (2(cos(t)sec(t))^1/2)

Answer 3

f(x) = sqrt(cos(t)) sec(t)

First off, sqrt(cos(t)) is the same as [cos(t)]^(1/2), and sec(t) is the same as 1/cos(t), so

f(x) = [cos(t)]^(1/2) / cos(t)

This can be reduced to

f(x) = [cos(t)]^(-1/2)

And we solve for the derivative using a combination of the power rule and chain rule.

f'(x) = (-1/2)[cos(t)]^(-3/2) [-sin(t)]

f'(x) = (1/2) [cos(t)]^(-3/2) [sin(t)]

Answer 4

In this problem, you would use the product rule, which states:

If

F(x) = uv

then

F ' (x) = uv' + vu'


In this case,

u = sqrt (cos t) = (cos t)^ (1/2)
u' = 1/ [ 2 sqrt (cost) ] * -sinx = 1/2 (cost) ^ (-1/2) * -sinx
= -sinx / ( 2 sqrt. (cos x) )


v = sec t
v' = sec t tan t

F ' ( t ) = [(sqrt (cos t))(sec t tan t)] + [(sec t)(-sinx / ( 2 sqrt. (cos x) ))]

Answer 5

this is the same as (cos(t)sec(t))^1/2
first use the power rule
1/2(cos(t)sec(t))^-1/2 times the derivative of cossec which we can find using the product rule:
-sin(t)sec(t)+ sec(t)sec(t)tan(t)
so you really have 1/2(cos(t)sec(t))^-1/2 times (-sin(t)sec(t)+sec(t)^2tan(t) which can be simplified to

-sin(t)sec(t)+sec(t)^2tan(t) all divided by (2(cos(t)sec(t))^1/2)

Answer 6

cos t sec t = 1
so
root(cos t sec t) = 1
constant
so derivative = 0

so you are right

Answer 7

d(sqrt cos(t)sec(t))/dt = d(sqrt cos(t)1/cos(t))/dt
= d(sqrt 1)/dt
= d(1)/dt
= 0

thank you