(\partial)/(\partial x)(2yzln(1+x^2))

Lösung

\frac{\partial}{\partial x} (2 yz ln (1 + x^{2}))

\frac{4 yz x}{1 + x ^{2}}

Schritte zur Lösung

\frac{\partial}{\partial x} (2 yz ln (1 + x^{2}))

Behandle

y, z

als Konstanten

Entferne die Konstante:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 yz \frac{\partial}{\partial x} (ln (1 + x^{2}))

Wende die Kettenregel an:

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

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

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

= 2 yz \frac{1}{1 + x ^{2}} \cdot 2 x

Vereinfache

2 yz \frac{1}{1 + x ^{2}} \cdot 2 x : \frac{4 yz x}{1 + x ^{2}}

= \frac{4 yz x}{1 + x ^{2}}

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