Differentiate. y = sec(𝜃) tan(𝜃)?

Let A=theta, then
y=secAtanA
=>
y'=secAtan^2(A)+sec^3(A)
=>
y'=secA[tan^2(A)+sec^2(A)]
=>
y'=secA[2sec^2(A)-1]...Show more

y = sec(θ).tan(θ) → recall: sec(x) = 1/cos(x)

y = [1/cos(θ)].tan(θ) → recall: tan(x) = sin(x)/cos(x)

y = [1/cos(θ)].[sin(θ)/cos(θ)]

y = sin(θ)/cos²(θ)

The function f looks like (u/v), so its derivative looks like: [(u'.v) - (v'.u)]/v² → where:

u = sin(θ) → u' = cos(θ)

v = cos²(θ) → v' = - 2.sin(θ).cos(θ)


y' = [(u'.v) - (v'.u)]/v²

y' = { [cos(θ) * cos²(θ)] - [- 2.sin(θ).cos(θ) * sin(θ)] } / [cos²(θ)]²

y' = [cos³(θ) + 2.sin²(θ).cos(θ)] / cos⁴(θ)

y' = [cos²(θ) + 2.sin²(θ)] / cos³(θ)

y' = [cos²(θ) + 2.{1 - cos²(θ)}] / cos³(θ)

y' = [cos²(θ) + 2 - 2.cos²(θ)] / cos³(θ)

y' = [2 - cos²(θ)]/cos³(θ) → to go further if you want it

y' = [2/cos³(θ)] - [cos²(θ)/cos³(θ)]

y' = 2.[1/cos(θ)]³ - [1/cos(θ)] → recall: sec(x) = 1/cos(x)

y' = 2.sec³(θ) - sec(θ)

y' = sec(θ).[2.sec²(θ) - 1]...Show more

Hint:  (f * g)' = ?...Show more

u = sec(t)
u' = sec(t) * tan(t)
v = tan(t)
v' = sec(t)^2

(u * v)' = u * v' + v * u'

sec(t) * sec(t)^2 + sec(t) * tan(t) * tan(t) =>
sec(t) * (sec(t)^2 + tan(t)^2) =>
sec(t) * (sec(t)^2 + sec(t)^2 - 1) =>
sec(t) * (2 * sec(t)^2 - 1) =>
2 * sec(t)^3 - sec(t)...Show more

y = sec (θ) tan (θ)
y' = (tan(θ) sec(θ))'...Show more