derivative ln|sec(θ)+tan(θ)|

解答

导数

ln ∣ sec (θ) + tan (θ) ∣

sec (θ)

求解步骤

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

使用连锁律:

\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 ) )})

化简

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

= sec (θ)

\int_{0}^{1} \frac{x ^{4} + 2}{x} d x

d er i v a t i v e \frac{s - 1}{s + 1}

\frac{\partial}{\partial x} (e^{y} e^{x})

\frac{\partial}{\partial x} (y e^{- 2 x})

l a pl a ce t r an s f or m - 2 t