derivative of 2x{z}(xcos(sin(x)))

Lösung

\frac{d}{d x} (2 x z (x) cos (sin (x)))

2 (cos (sin (x)) z (x) + x (cos (sin (x)) \frac{d}{d x} (z (x)) - sin (sin (x)) cos (x) z (x)))

Schritte zur Lösung

\frac{d}{d x} (2 x z (x) cos (sin (x)))

Entferne die Konstante:

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

= 2 \frac{d}{d x} (x z (x) cos (sin (x)))

Wende die Produktregel an:

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

f = x, g = z (x) cos (sin (x))

= 2 (\frac{d x}{d x} z (x) cos (sin (x)) + \frac{d}{d x} (z (x) cos (sin (x))) x)

\frac{d x}{d x} = 1

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

= 2 (1 \cdot z (x) cos (sin (x)) + (cos (sin (x)) \frac{d}{d x} (z (x)) - sin (sin (x)) cos (x) z (x)) x)

Vereinfache

= 2 (cos (sin (x)) z (x) + x (cos (sin (x)) \frac{d}{d x} (z (x)) - sin (sin (x)) cos (x) z (x)))

\frac{d x}{d t} = 3 (x^{2} + 1), x (\frac{π}{4}) = 1

d er i v a t i v e f (x) = \frac{2 x - 3}{4 - x}

\frac{\partial}{\partial x} (\frac{x}{1 + x ^{2}})

a re a y (x + 4) = 1, x = 0, y = 0, x = 4

\int (ln (16 x))^{2} d x