derivative of e^{xy}cos(x+y)

Lösung

\frac{d}{d x} (e^{x y} cos (x + y))

e^{x y} cos (x + y) (y + x \frac{d y}{d x}) - e^{x y} sin (x + y) (1 + \frac{d y}{d x})

Schritte zur Lösung

\frac{d}{d x} (e^{x y} cos (x + y))

Wende die Produktregel an:

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

f = e^{x y}, g = cos (x + y)

= \frac{d}{d x} (e^{x y}) cos (x + y) + \frac{d}{d x} (cos (x + y)) e^{x y}

\frac{d}{d x} (e^{x y}) = e^{x y} (y + x \frac{d y}{d x})

\frac{d}{d x} (cos (x + y)) = - sin (x + y) (1 + \frac{d y}{d x})

= e^{x y} (y + x \frac{d y}{d x}) cos (x + y) + (- sin (x + y) (1 + \frac{d y}{d x})) e^{x y}

Vereinfache

= e^{x y} cos (x + y) (y + x \frac{d y}{d x}) - e^{x y} sin (x + y) (1 + \frac{d y}{d x})

y^{^{'}} - \frac{3}{x ^{2}} y = \frac{1}{x ^{2}}

y^{^{'}} = \frac{t}{t ^{2} y + y}

n = 1 \sum \infty n e^{- \frac{n}{8}}

y^{^{''}} + 20 y^{^{'}} = 0, y (0) = - 5, y^{^{'}} (0) = 20

\frac{d ^{2}}{d x ^{2}} (x^{\frac{1}{11}})