(\partial)/(\partial y)(1/(t(y^2-9)))

Lösung

\frac{\partial}{\partial y} (\frac{1}{t ( y ^{2} - 9 )})

- \frac{2 y}{t ( y ^{2} - 9 ) ^{2}}

Schritte zur Lösung

\frac{\partial}{\partial y} (\frac{1}{t ( y ^{2} - 9 )})

Behandle

t

als Konstante

Entferne die Konstante:

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

= \frac{1}{t} \frac{\partial}{\partial y} (\frac{1}{y ^{2} - 9})

Wende Exponentenregel an:

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

= \frac{1}{t} \frac{\partial}{\partial y} ((y^{2} - 9)^{- 1})

Wende die Kettenregel an:

- \frac{1}{( y ^{2} - 9 ) ^{2}} \frac{\partial}{\partial y} (y^{2} - 9)

= - \frac{1}{( y ^{2} - 9 ) ^{2}} \frac{\partial}{\partial y} (y^{2} - 9)

\frac{\partial}{\partial y} (y^{2} - 9) = 2 y

= \frac{1}{t} (- \frac{1}{( y ^{2} - 9 ) ^{2}} \cdot 2 y)

Vereinfache

\frac{1}{t} (- \frac{1}{( y ^{2} - 9 ) ^{2}} \cdot 2 y) : - \frac{2 y}{t ( y ^{2} - 9 ) ^{2}}

= - \frac{2 y}{t ( y ^{2} - 9 ) ^{2}}

a re a f (x) = \frac{x}{9 - x ^{2}}, g (y) = 0, y = 2

\frac{d}{d x} (arcsin (2 x - 1))

y^{^{''}} - 2 y^{^{'}} - 3 y = 3 e^{2 t}, y (0) = 1, y^{^{'}} (0) = 0

x \to \infty lim (\frac{1}{3})

\frac{d y}{d x} + 2 y - 8 e^{- x} = 0, y (0) = 7