(\partial)/(\partial x)(3cos(3x+4y))

Lösung

\frac{\partial}{\partial x} (3 cos (3 x + 4 y))

- 9 sin (3 x + 4 y)

Schritte zur Lösung

\frac{\partial}{\partial x} (3 cos (3 x + 4 y))

Behandle

y

als Konstante

Entferne die Konstante:

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

= 3 \frac{\partial}{\partial x} (cos (3 x + 4 y))

Wende die Kettenregel an:

- sin (3 x + 4 y) \frac{\partial}{\partial x} (3 x + 4 y)

= - sin (3 x + 4 y) \frac{\partial}{\partial x} (3 x + 4 y)

\frac{\partial}{\partial x} (3 x + 4 y) = 3

= 3 (- sin (3 x + 4 y) \cdot 3)

Vereinfache

= - 9 sin (3 x + 4 y)

\frac{d y}{d t} = \frac{t}{y - 2}, y (- 1) = 0

\int sin (x) sin (2 x) cos (3 x) d x

\frac{\partial}{\partial x} (\frac{- x y}{x ^{2} + y ^{2}})

x \to 3 + lim (\frac{3 - x}{x ^{2} - 9})

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