(\partial)/(\partial x)(y/(ln(x-1)))

Lösung

\frac{\partial}{\partial x} (\frac{y}{ln ( x - 1 )})

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

Schritte zur Lösung

\frac{\partial}{\partial x} (\frac{y}{ln ( x - 1 )})

Behandle

y

als Konstante

Entferne die Konstante:

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

= y \frac{\partial}{\partial x} (\frac{1}{ln ( x - 1 )})

Wende Exponentenregel an:

\frac{1}{a} = a^{- 1}

= y \frac{\partial}{\partial x} (ln^{- 1} (x - 1))

Wende die Kettenregel an:

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

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

\frac{\partial}{\partial x} (ln (x - 1)) = \frac{1}{x - 1}

= y (- \frac{1}{( ln ( x - 1 ) ) ^{2}} \cdot \frac{1}{x - 1})

Vereinfache

y (- \frac{1}{ln ^{2} ( x - 1 )} \cdot \frac{1}{x - 1}) : - \frac{y}{ln ^{2} ( x - 1 ) ( x - 1 )}

= - \frac{y}{ln ^{2} ( x - 1 ) ( x - 1 )}

\frac{d y}{d t} + a y = 0

2 y^{^{'}} + x y^{^{''}} = 0

\frac{\partial}{\partial z} (2 x)

\int 4 x sec^{2} (2 x) d x

\int x - 4 x d x