derivative of (sqrt(x^2+y^2-9)/x)

Lösung

\frac{d}{d x} (\frac{x ^{2} + y ^{2} - 9}{x})

\frac{x y \frac{d y}{d x} - y ^{2} + 9}{x ^{2} x ^{2} + y ^{2} - 9}

Schritte zur Lösung

\frac{d}{d x} (\frac{x ^{2} + y ^{2} - 9}{x})

Wende die Quotientenregel an:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

= \frac{\frac{x + y \frac{d y}{d x}}{x ^{2} + y ^{2} - 9} x - 1 \cdot x ^{2} + y ^{2} - 9}{x ^{2}}

Vereinfache

\frac{\frac{x + y \frac{d y}{d x}}{x ^{2} + y ^{2} - 9} x - 1 \cdot x ^{2} + y ^{2} - 9}{x ^{2}} : \frac{x y \frac{d y}{d x} - y ^{2} + 9}{x ^{2} x ^{2} + y ^{2} - 9}

= \frac{x y \frac{d y}{d x} - y ^{2} + 9}{x ^{2} x ^{2} + y ^{2} - 9}

n = 1 \sum \infty \frac{- 1 ^{n + 1}}{n}

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

\int_{0}^{1} \frac{6}{1 + t ^{2}} d t

\frac{d}{d x} (\frac{cos ( x )}{x} + \frac{x}{cos ( x )})

d er i v a t i v e \frac{1}{4 x ^{2} + 1}