(\partial)/(\partial x)(ln(7x^2+y^2+6))

Lösung

\frac{\partial}{\partial x} (ln (7 x^{2} + y^{2} + 6))

\frac{14 x}{7 x ^{2} + y ^{2} + 6}

Schritte zur Lösung

\frac{\partial}{\partial x} (ln (7 x^{2} + y^{2} + 6))

Behandle

y

als Konstante

Wende die Kettenregel an:

\frac{1}{7 x ^{2} + y ^{2} + 6} \frac{\partial}{\partial x} (7 x^{2} + y^{2} + 6)

= \frac{1}{7 x ^{2} + y ^{2} + 6} \frac{\partial}{\partial x} (7 x^{2} + y^{2} + 6)

\frac{\partial}{\partial x} (7 x^{2} + y^{2} + 6) = 14 x

= \frac{1}{7 x ^{2} + y ^{2} + 6} \cdot 14 x

Vereinfache

\frac{1}{7 x ^{2} + y ^{2} + 6} \cdot 14 x : \frac{14 x}{7 x ^{2} + y ^{2} + 6}

= \frac{14 x}{7 x ^{2} + y ^{2} + 6}

x y^{^{'}} + 2 y = - 2 e^{x}

x \to - 3 lim (\frac{7}{x + 3})

y^{^{'}} = t^{2} - y

\frac{d}{d x} (arccos (\frac{3}{x}))

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