derivative of sin^{ln(x}(3x))

Lösung

\frac{d}{d x} (sin^{l n (x)} (3 x))

x^{l n (s i n (3 x))} (\frac{ln ( sin ( 3 x ) )}{x} + 3 cot (3 x) ln (x))

Schritte zur Lösung

\frac{d}{d x} (sin^{l n (x)} (3 x))

Wende Exponentenregel an:

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

sin^{l n (x)} (3 x) = e^{l n (x) l n (s i n (3 x))}

= \frac{d}{d x} (e^{l n (x) l n (s i n (3 x))})

Wende die Kettenregel an:

e^{l n (x) l n (s i n (3 x))} \frac{d}{d x} (ln (x) ln (sin (3 x)))

= e^{l n (x) l n (s i n (3 x))} \frac{d}{d x} (ln (x) ln (sin (3 x)))

\frac{d}{d x} (ln (x) ln (sin (3 x))) = \frac{ln ( sin ( 3 x ) )}{x} + 3 cot (3 x) ln (x)

= e^{l n (x) l n (s i n (3 x))} (\frac{ln ( sin ( 3 x ) )}{x} + 3 cot (3 x) ln (x))

e^{l n (x) l n (s i n (3 x))} = x^{l n (s i n (3 x))}

= x^{l n (s i n (3 x))} (\frac{ln ( sin ( 3 x ) )}{x} + 3 cot (3 x) ln (x))

\int sin (a t) d t

\frac{d}{d x} (e^{3 x} + 10 e^{2 x})

t^{2} y^{^{'}} + 3 t y + 2 t^{3} = 0

(e^{y} + 6 x) d x + (x e^{y}) d y = 0

\int e^{l n (2 t)} d t