derivative of arccos(x^2+y^2-4)

Lösung

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

- \frac{2 x + 2 y \frac{d y}{d x}}{- x ^{4} + 8 x ^{2} - 2 x ^{2} y ^{2} + 8 y ^{2} - y ^{4} - 15}

Schritte zur Lösung

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

Wende die Kettenregel an:

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

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

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

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

Vereinfache

- \frac{1}{1 - ( x ^{2} + y ^{2} - 4 ) ^{2}} (2 x + 2 y \frac{d y}{d x}) : - \frac{2 x + 2 y \frac{d y}{d x}}{- x ^{4} + 8 x ^{2} - 2 x ^{2} y ^{2} + 8 y ^{2} - y ^{4} - 15}

= - \frac{2 x + 2 y \frac{d y}{d x}}{- x ^{4} + 8 x ^{2} - 2 x ^{2} y ^{2} + 8 y ^{2} - y ^{4} - 15}

y = c e^{x}

y^{^{''}} - 2 y^{^{'}} + 8 y = 0

a re a - x^{2} + 3, 3 x^{2} - 9 x

\int_{0}^{a} x^{2} a^{2} - x^{2} d x

\int (3 x^{2} - 4 x + 5) d x