(dy)/(dx)=(2x)/((1+x^2))

Lösung

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

y = ln (1 + x^{2}) + c_{1}

Schritte zur Lösung

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

Ersetze

\frac{d y}{d x}

mit

y^{'} (x)

y^{^{'}} (x) = \frac{2 x}{( 1 + x ^{2} )}

Wenn

f^{^{'}} (x) = g (x)

dann

f (x) = \int g (x) d x

y = \int \frac{2 x}{1 + x ^{2}} d x

\int \frac{2 x}{1 + x ^{2}} d x = ln (1 + x^{2}) + c_{1}

y = ln (1 + x^{2}) + c_{1}

t an g e n t y = \frac{5}{4 x + 1}

\frac{d}{d x} (\frac{3 - 17 x ^{2}}{3 x ^{3}})

d er i v a t i v e 5 e^{x} tan (x)

\int \frac{4 x ^{2} + 3 x - 1}{x ^{3} - x ^{2}} d x

\frac{\partial}{\partial y} (\frac{y}{x + z})