(\partial)/(\partial x)((5x)/(x^2+y^2))

Lösung

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

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

Schritte zur Lösung

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

Behandle

y

als Konstante

Entferne die Konstante:

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

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

Wende die Quotientenregel an:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

= 5 \cdot \frac{1 \cdot ( x ^{2} + y ^{2} ) - 2 xx}{( x ^{2} + y ^{2} ) ^{2}}

Vereinfache

5 \cdot \frac{1 \cdot ( x ^{2} + y ^{2} ) - 2 xx}{( x ^{2} + y ^{2} ) ^{2}} : \frac{5 ( - x ^{2} + y ^{2} )}{( x ^{2} + y ^{2} ) ^{2}}

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

f (x) = 2^{x} ln (2)

n = 0 \sum \infty \frac{1}{n + 15}

\int_{0}^{0.1} f (10 t) d t

\int (ln (x))^{0} d x

\int \frac{1}{x ^{2} 5 - x ^{2}} d x