derivative of sqrt(1/4 x^2-2x-y)

Lösung

\frac{d}{d x} (\frac{1}{4} x^{2} - 2 x - y)

\frac{x - 4 - 2 \frac{d y}{d x}}{2 x ^{2} - 8 x - 4 y}

Schritte zur Lösung

\frac{d}{d x} (\frac{1}{4} x^{2} - 2 x - y)

Wende die Kettenregel an:

\frac{1}{2 \frac{1}{4} x ^{2} - 2 x - y} \frac{d}{d x} (\frac{1}{4} x^{2} - 2 x - y)

= \frac{1}{2 \frac{1}{4} x ^{2} - 2 x - y} \frac{d}{d x} (\frac{1}{4} x^{2} - 2 x - y)

\frac{d}{d x} (\frac{1}{4} x^{2} - 2 x - y) = \frac{x}{2} - 2 - \frac{d y}{d x}

= \frac{1}{2 \frac{1}{4} x ^{2} - 2 x - y} (\frac{x}{2} - 2 - \frac{d y}{d x})

Vereinfache

\frac{1}{2 \frac{1}{4} x ^{2} - 2 x - y} (\frac{x}{2} - 2 - \frac{d y}{d x}) : \frac{x - 4 - 2 \frac{d y}{d x}}{2 x ^{2} - 8 x - 4 y}

= \frac{x - 4 - 2 \frac{d y}{d x}}{2 x ^{2} - 8 x - 4 y}

\frac{\partial}{\partial y} (ln (x y))

\int x^{2} 3 - 2 x d x

\int \frac{1}{( a ^{2} + v ^{2} )} d v

y^{^{'}} + tan (x) y - sin (x) = 0

\int \frac{1}{x} + 2 x d x