derivative of (tan(x)^{sec(x)})

Lösung

\frac{d}{d x} ((tan (x))^{s e c (x)})

tan^{s e c (x)} (x) (sec (x) tan (x) ln (tan (x)) + 2 csc (2 x) sec (x))

Schritte zur Lösung

\frac{d}{d x} ((tan (x))^{s e c (x)})

Wende Exponentenregel an:

a^{b} = e^{b l n (a)}

(tan (x))^{s e c (x)} = e^{s e c (x) l n (t a n (x))}

= \frac{d}{d x} (e^{s e c (x) l n (t a n (x))})

Wende die Kettenregel an:

e^{s e c (x) l n (t a n (x))} \frac{d}{d x} (sec (x) ln (tan (x)))

= e^{s e c (x) l n (t a n (x))} \frac{d}{d x} (sec (x) ln (tan (x)))

\frac{d}{d x} (sec (x) ln (tan (x))) = sec (x) tan (x) ln (tan (x)) + 2 csc (2 x) sec (x)

= e^{s e c (x) l n (t a n (x))} (sec (x) tan (x) ln (tan (x)) + 2 csc (2 x) sec (x))

e^{s e c (x) l n (t a n (x))} = tan^{s e c (x)} (x)

= tan^{s e c (x)} (x) (sec (x) tan (x) ln (tan (x)) + 2 csc (2 x) sec (x))

\int tan^{2} (y) + 1 d y

\frac{\partial}{\partial x} (2 x^{2} + y^{2} - 4 y + 4)

a re a 0.5 (e^{x} + e^{- x}), 0 \leq x \leq ln (2)

\frac{d}{d x} (\frac{lo g _{10} ( x )}{ln ( x )} + 2 x)

y^{^{''}} + 5 y^{^{'}} + 6 y = 2 e^{x}