derivative f(x)=x^3arcsin(x^2e^x)

Lösung

ableitung von

f (x) = x^{3} arcsin (x^{2} e^{x})

3 x^{2} arcsin (e^{x} x^{2}) + \frac{x ^{3} ( 2 e ^{x} x + e ^{x} x ^{2} )}{1 - e ^{2 x} x ^{4}}

Schritte zur Lösung

\frac{d}{d x} (x^{3} arcsin (x^{2} e^{x}))

Wende die Produktregel an:

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

f = x^{3}, g = arcsin (x^{2} e^{x})

= \frac{d}{d x} (x^{3}) arcsin (x^{2} e^{x}) + \frac{d}{d x} (arcsin (x^{2} e^{x})) x^{3}

\frac{d}{d x} (x^{3}) = 3 x^{2}

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

= 3 x^{2} arcsin (x^{2} e^{x}) + \frac{2 e ^{x} x + e ^{x} x ^{2}}{1 - e ^{2 x} x ^{4}} x^{3}

Multipliziere Brüche:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= 3 x^{2} arcsin (e^{x} x^{2}) + \frac{x ^{3} ( 2 e ^{x} x + e ^{x} x ^{2} )}{1 - e ^{2 x} x ^{4}}

in v erse l a pl a ce \frac{s - 2}{s ^{2} + 5 s + 6}

d er i v a t i v e \frac{x ^{3}}{4} + \frac{1}{3 x}

\frac{\partial}{\partial t} (3 t)

\frac{d v}{d t} = - 2 v - 32

y^{^{''}} = - 16 y