derivative y=sqrt(1-x^2)arcsin(x)

Lösung

ableitung von

y = 1 - x^{2} arcsin (x)

- \frac{x arcsin ( x )}{1 - x ^{2}} + 1

Schritte zur Lösung

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

Wende die Produktregel an:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = 1 - x^{2}, g = arcsin (x)

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

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

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

= (- \frac{x}{1 - x ^{2}}) arcsin (x) + \frac{1}{1 - x ^{2}} 1 - x^{2}

Vereinfache

(- \frac{x}{1 - x ^{2}}) arcsin (x) + \frac{1}{1 - x ^{2}} 1 - x^{2} : - \frac{x arcsin ( x )}{1 - x ^{2}} + 1

= - \frac{x arcsin ( x )}{1 - x ^{2}} + 1

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

d er i v a t i v e \frac{- x ^{2} - 8}{( x ^{2} - 8 ) ^{2}}

\int_{\frac{1}{3}}^{5} 14 x ln (3 x) d x

in v erse l a pl a ce \frac{( e ^{- s} )}{s ^{2} - 6 s + 9}

y^{^{'}} = \frac{2 y}{t} + t