d/(dθ)(1/4 sin^2(2θ))

Lösung

\frac{d}{d θ} (\frac{1}{4} sin^{2} (2 θ))

\frac{sin ( 4 θ )}{2}

Schritte zur Lösung

\frac{d}{d θ} (\frac{1}{4} sin^{2} (2 θ))

Entferne die Konstante:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{4} \frac{d}{d θ} (sin^{2} (2 θ))

Wende die Kettenregel an:

2 sin (2 θ) \frac{d}{d θ} (sin (2 θ))

= 2 sin (2 θ) \frac{d}{d θ} (sin (2 θ))

\frac{d}{d θ} (sin (2 θ)) = cos (2 θ) \cdot 2

= \frac{1}{4} \cdot 2 sin (2 θ) cos (2 θ) \cdot 2

Vereinfache

\frac{1}{4} \cdot 2 sin (2 θ) cos (2 θ) \cdot 2 : \frac{sin ( 4 θ )}{2}

= \frac{sin ( 4 θ )}{2}

n = 0 \sum \infty e^{- \frac{3 n}{2}}

\int 4 x^{2} + 2 + \frac{5}{x ^{2} + 1} d x

\frac{\partial}{\partial t} (3 arcsin (t))

t an g e n t (x^{2} + y^{2})^{2} = 4 x^{2} y, (- 1, 1)

\int_{1}^{10} \frac{x ^{2} - 1}{x} d x