implicit (dy)/(dx),ax^3-3b^2xy+cy(x)^3=1

Lösung

implizite ableitung

\frac{d y}{d x}, a x^{3} - 3 b^{2} x y + cy (x)^{3} = 1

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

Schritte zur Lösung

a x^{3} - 3 b^{2} x y + cy x^{3} = 1

Behandle

y

wie

y (x)

Beide Seiten differenzieren:

3 a x^{2} - 3 b^{2} (y + x \frac{d y}{d x}) + c (\frac{d y}{d x} x^{3} + 3 x^{2} y) = 0

Isoliere

\frac{d y}{d x} : \frac{d y}{d x} = \frac{- 3 a x ^{2} + 3 b ^{2} y - 3 c x ^{2} y}{x ( c x ^{2} - 3 b ^{2} )}

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

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

(x^{2} + 1) y^{^{'}} + 8 x (y - 1) = 0

d er i v a t i v e x^{4} - 8 x^{3}

\frac{\partial}{\partial x} (\frac{3}{3 x + 2 y})

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