derivative of arcsin(3x+y)

Lösung

\frac{d}{d x} (arcsin (3 x + y))

\frac{3 + \frac{d y}{d x}}{1 - 9 x ^{2} - 6 x y - y ^{2}}

Schritte zur Lösung

\frac{d}{d x} (arcsin (3 x + y))

Wende die Kettenregel an:

\frac{1}{1 - ( 3 x + y ) ^{2}} \frac{d}{d x} (3 x + y)

= \frac{1}{1 - ( 3 x + y ) ^{2}} \frac{d}{d x} (3 x + y)

\frac{d}{d x} (3 x + y) = 3 + \frac{d y}{d x}

= \frac{1}{1 - ( 3 x + y ) ^{2}} (3 + \frac{d y}{d x})

Vereinfache

\frac{1}{1 - ( 3 x + y ) ^{2}} (3 + \frac{d y}{d x}) : \frac{3 + \frac{d y}{d x}}{1 - 9 x ^{2} - 6 x y - y ^{2}}

= \frac{3 + \frac{d y}{d x}}{1 - 9 x ^{2} - 6 x y - y ^{2}}

\frac{d}{d x} (ln (5 x + 3))

\int 4 e^{- x} (e^{- 3 x} + 1) d x

\frac{d}{d x} (\frac{9 x}{e ^{x}})

d er i v a t i v e f (x) = arccos (3 x)

\frac{d}{d x} (arctan (x - 1 + x^{2}))