derivative of ln(x^2+y^2+1)

Lösung

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

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

Schritte zur Lösung

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

Wende die Kettenregel an:

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

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

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

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

Vereinfache

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

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

x \to \infty + lim (\frac{( 5 x ^{3} + 7 x ^{2} - 8 x + 9 )}{11 x ^{2} + 7 x - 6})

\int sin^{4} (7 x) d x

in v erse l a pl a ce \frac{1}{( s - 5 ) ^{2}}

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

y^{^{''}} + 4 y^{^{'}} + 13 y = 0