derivative 1/x sin^3(6x)

Lösung

ableitung von

\frac{1}{x} sin^{3} (6 x)

- \frac{sin ^{3} ( 6 x )}{x ^{2}} + \frac{18 sin ^{2} ( 6 x ) cos ( 6 x )}{x}

Schritte zur Lösung

\frac{d}{d x} (\frac{1}{x} sin^{3} (6 x))

Wende die Produktregel an:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = \frac{1}{x}, g = sin^{3} (6 x)

= \frac{d}{d x} (\frac{1}{x}) sin^{3} (6 x) + \frac{d}{d x} (sin^{3} (6 x)) \frac{1}{x}

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

\frac{d}{d x} (sin^{3} (6 x)) = 18 sin^{2} (6 x) cos (6 x)

= (- \frac{1}{x ^{2}}) sin^{3} (6 x) + 18 sin^{2} (6 x) cos (6 x) \frac{1}{x}

Vereinfache

(- \frac{1}{x ^{2}}) sin^{3} (6 x) + 18 sin^{2} (6 x) cos (6 x) \frac{1}{x} : - \frac{sin ^{3} ( 6 x )}{x ^{2}} + \frac{18 sin ^{2} ( 6 x ) cos ( 6 x )}{x}

= - \frac{sin ^{3} ( 6 x )}{x ^{2}} + \frac{18 sin ^{2} ( 6 x ) cos ( 6 x )}{x}

\int - 0.6 x^{2} + 80 d x

d er i v a t i v e 1 + 2 x

\int \frac{1}{t ^{2} - 1} d t

\frac{d}{d x} (x^{2} + \frac{y ^{2}}{9}) = 1

l a pl a ce t r an s f or m t^{\frac{3}{2}}