Calculus Help?

Question

Find g'(t)

g(t)= sec(t) cos^2(t) tan(t)

Answer 1

To solve this, remember that sec(t) = 1/cos(t), this comes up a lot. This looks complex, but it actually ends up simpler than it originally appears.

g(t)= sec(t) cos^2(t) tan(t)

Since sec (t) = 1/cos(t), sec(t) * cos^2(t) = cos^2(t)/cos(t) = cos(t).

g(t)= cos(t) tan(t)

Product rule: (fg)' = gf' + fg'
f = cos(t)
f' = -sin(t)
g = tan(t)
g' = sec^2(t)

g'(t) = -tan(t)sin(t) + cos(t)sec^2(t)

Again, sec(t) = 1/cos(t), so cos(t)sec^2(t) = sec(t)

g'(t) = sec(t) - sin(t)tan(t)

We can continue:
tan(t) = sin(t)/cos(t) = sin(t)sec(t)
g'(t) = sec(t) - sin(t) * sin(t)sec(t)
g'(t) = sec(t) - sin^2(t)sec(t)
g'(t) = sec(t) * (1 - sin^2)

Trig identity: sin^2(t) + cos^2(t) = 1, so cos^2 = 1 - sin^2

g'(t) = sec(t) * cos^2(t)
g(t) = cos(t)

Answer 2

sec(t) = 1/cos(t)
tan(t) = sin(t)/cos(t)
g(t) = sec(t) * cos^2(t) * tan(t) = 1/cos(t) * cos^2(t) * sin(t)/cos(t) = sin(t)
g'(t) = cos(t)

It's the first time in 35 years when I use sec(t) !!!