derivative of-x(2x-y^3)

Lösung

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

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

Schritte zur Lösung

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

Entferne die Konstante:

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

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

Wende die Produktregel an:

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

f = x, g = 2 x - y^{3}

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

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

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

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

Vereinfache

- (1 \cdot (2 x - y^{3}) + (2 - 3 y^{2} \frac{d y}{d x}) x) : - 4 x + 3 x y^{2} \frac{d y}{d x} + y^{3}

= - 4 x + 3 x y^{2} \frac{d y}{d x} + y^{3}

\int (4 + x) d x

t a y l or e^{\frac{1}{θ}}

n = 0 \sum \infty \frac{3}{n n}

\int \frac{sin ( 2 x )}{1 + cos ^{2} ( 2 x )} d x

y^{^{'}} = y (2 + y) t, y (0) = 1