What is the derivative of ln|sec(θ)+tan(θ)| ?

The derivative of ln|sec(θ)+tan(θ)| is sec(θ)

What is the first derivative of ln|sec(θ)+tan(θ)| ?

The first derivative of ln|sec(θ)+tan(θ)| is sec(θ)

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derivative of

ln ∣ sec (θ) + tan (θ) ∣

sec (θ)

Solution steps

\frac{d}{d θ} (ln ∣ sec (θ) + tan (θ) ∣)

Apply the chain rule:

\frac{1}{∣ sec ( θ ) + tan ( θ ) ∣} \frac{d}{d θ} (∣ sec (θ) + tan (θ) ∣)

= \frac{1}{∣ sec ( θ ) + tan ( θ ) ∣} \frac{d}{d θ} (∣ sec (θ) + tan (θ) ∣)

\frac{d}{d θ} (∣ sec (θ) + tan (θ) ∣) = - \frac{sec ( θ ) + tan ( θ )}{∣ sec ( θ ) + tan ( θ ) ∣ ( - 1 + sin ( θ ) )}

= \frac{1}{∣ sec ( θ ) + tan ( θ ) ∣} (- \frac{sec ( θ ) + tan ( θ )}{∣ sec ( θ ) + tan ( θ ) ∣ ( - 1 + sin ( θ ) )})

Simplify

\frac{1}{∣ sec ( θ ) + tan ( θ ) ∣} (- \frac{sec ( θ ) + tan ( θ )}{∣ sec ( θ ) + tan ( θ ) ∣ ( - 1 + sin ( θ ) )}) : sec (θ)

= sec (θ)

What is the derivative of ln|sec(θ)+tan(θ)| ?
The derivative of ln|sec(θ)+tan(θ)| is sec(θ)
What is the first derivative of ln|sec(θ)+tan(θ)| ?
The first derivative of ln|sec(θ)+tan(θ)| is sec(θ)