derivative of arcsech(sin(x))

Lösung

\frac{d}{d x} (arcsech (sin (x)))

- \frac{cot ( x )}{sin ( x ) + 1 - sin ( x ) + 1}

Schritte zur Lösung

\frac{d}{d x} (arcsech (sin (x)))

Wende die Kettenregel an:

- \frac{1}{sin ( x ) sin ( x ) + 1 - sin ( x ) + 1} \frac{d}{d x} (sin (x))

= - \frac{1}{sin ( x ) sin ( x ) + 1 - sin ( x ) + 1} \frac{d}{d x} (sin (x))

\frac{d}{d x} (sin (x)) = cos (x)

= - \frac{1}{sin ( x ) sin ( x ) + 1 - sin ( x ) + 1} cos (x)

Vereinfache

- \frac{1}{sin ( x ) sin ( x ) + 1 - sin ( x ) + 1} cos (x) : - \frac{cot ( x )}{sin ( x ) + 1 - sin ( x ) + 1}

= - \frac{cot ( x )}{sin ( x ) + 1 - sin ( x ) + 1}

\int e^{x} x^{4} d x

\frac{d}{d θ} (a^{2} \cdot sin^{2} (θ) + b^{2} \cdot cos^{2} (θ))

\frac{d y}{d x} = - 6 (y^{2} - y)

d er i v a t i v e \frac{2 ln ( 2 x )}{2 - 3 x}

\frac{d}{d x} ((x^{2} + y^{2})^{\frac{2}{3}})