(\partial)/(\partial x)(2sqrt(x^2+2y^2))

Lösung

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

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

Schritte zur Lösung

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

Behandle

y

als Konstante

Entferne die Konstante:

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

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

Wende die Kettenregel an:

\frac{1}{2 x ^{2} + 2 y ^{2}} \frac{\partial}{\partial x} (x^{2} + 2 y^{2})

= \frac{1}{2 x ^{2} + 2 y ^{2}} \frac{\partial}{\partial x} (x^{2} + 2 y^{2})

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

= 2 \cdot \frac{1}{2 x ^{2} + 2 y ^{2}} \cdot 2 x

Vereinfache

2 \cdot \frac{1}{2 x ^{2} + 2 y ^{2}} \cdot 2 x : \frac{2 x}{x ^{2} + 2 y ^{2}}

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

\frac{d y}{d x} = \frac{y ^{2} - 1}{x ^{2} - 1}, y (9) = 9

\frac{d}{d y} ((\frac{1}{y ^{2}} - \frac{3}{y ^{4}}) (y + 9 y^{3}))

d er i v a t i v e 2 csc (1 - 3 x)

t an g e n t f (x) = (2 x + 3)^{5}, at x = - 1

\frac{d}{d x} (4 x - \frac{4}{x})