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

Lösung

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

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

Schritte zur Lösung

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

Behandle

y

als Konstante

Wende Exponentenregel an:

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

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

Wende die Kettenregel an:

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

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

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

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

Vereinfache

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

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

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

\int 3 y e^{y^{3}} d y

f (x) = ln (x - 4)

\int - cos^{2} (t) d t

\frac{\partial}{\partial x} (2^{x^{2} - y^{5}})