(\partial)/(\partial x)(sqrt(33-3x^2-3y^2))

Lösung

\frac{\partial}{\partial x} (33 - 3 x^{2} - 3 y^{2})

- \frac{3 x}{33 - 3 x ^{2} - 3 y ^{2}}

Schritte zur Lösung

\frac{\partial}{\partial x} (33 - 3 x^{2} - 3 y^{2})

Behandle

y

als Konstante

Wende die Kettenregel an:

\frac{1}{2 33 - 3 x ^{2} - 3 y ^{2}} \frac{\partial}{\partial x} (33 - 3 x^{2} - 3 y^{2})

= \frac{1}{2 33 - 3 x ^{2} - 3 y ^{2}} \frac{\partial}{\partial x} (33 - 3 x^{2} - 3 y^{2})

\frac{\partial}{\partial x} (33 - 3 x^{2} - 3 y^{2}) = - 6 x

= \frac{1}{2 33 - 3 x ^{2} - 3 y ^{2}} (- 6 x)

Vereinfache

\frac{1}{2 33 - 3 x ^{2} - 3 y ^{2}} (- 6 x) : - \frac{3 x}{33 - 3 x ^{2} - 3 y ^{2}}

= - \frac{3 x}{33 - 3 x ^{2} - 3 y ^{2}}

t an g e n t f (x) = \frac{x}{x ^{2} - 1}, at x = 2

d er i v a t i v e (5 x^{6} + 2 x^{3})^{4}

d er i v a t i v e f (x) = (5 + x) (12 - x) - 2.2 x

\frac{d}{d x} (sec^{2} (π x))

\frac{d}{d x} (\frac{x ^{4}}{4 - x ^{3}})