erweitern (dx)/d (x^2+y^2=y^3+2x-y)

Lösung

erweitern

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

\frac{x ^{2} d x}{d} + \frac{y ^{2} d x}{d} = y^{3} + \frac{2 x d x}{d} - \frac{y d x}{d}

Schritte zur Lösung

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

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

Setze Klammern

\frac{d x}{d} x^{2} + \frac{d x}{d} y^{2} = y^{3} + \frac{d x}{d} \cdot 2 x + \frac{d x}{d} (- y)

Wende Minus-Plus Regeln an

+ (- a) = - a

\frac{d x}{d} x^{2} + \frac{d x}{d} y^{2} = y^{3} + 2 \cdot \frac{d x}{d} x - \frac{d x}{d} y

Vereinfache

\frac{d x}{d} x^{2} + \frac{d x}{d} y^{2} = y^{3} + 2 \cdot \frac{d x}{d} x - \frac{d x}{d} y : \frac{x ^{2} d x}{d} + \frac{y ^{2} d x}{d} = y^{3} + \frac{2 x d x}{d} - \frac{y d x}{d}

\frac{x ^{2} d x}{d} + \frac{y ^{2} d x}{d} = y^{3} + \frac{2 x d x}{d} - \frac{y d x}{d}

e x p an d (- b sin (t) + b cos (t) + 1 - b)^{3}

e x p an d e^{x} (c_{1} cos (2 x) + c_{2} sin (2 x))

e x p an d t (e^{t} + e^{5 t})^{2}

e x p an d - \frac{2}{( x - 1 ) ^{3}}

e x p an d - \frac{V ( e ^{- a t} - 1 )}{a}