(\partial)/(\partial y)(3x^2+4xy+y^2)

Lösung

\frac{\partial}{\partial y} (3 x^{2} + 4 x y + y^{2})

4 x + 2 y

Schritte zur Lösung

\frac{\partial}{\partial y} (3 x^{2} + 4 x y + y^{2})

Behandle

x

als Konstante

Wende die Summen-/Differenzregel an:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{\partial}{\partial y} (3 x^{2}) + \frac{\partial}{\partial y} (4 x y) + \frac{\partial}{\partial y} (y^{2})

\frac{\partial}{\partial y} (3 x^{2}) = 0

\frac{\partial}{\partial y} (4 x y) = 4 x

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

= 0 + 4 x + 2 y

Vereinfache

= 4 x + 2 y

\frac{\partial}{\partial x} (6 x y)

y^{^{''}} + y = 0, y (\frac{π}{3}) = 0, y^{^{'}} (\frac{π}{3}) = 4

\frac{d}{d x} ((tan (\frac{1}{x}))^{s e c (\frac{1}{x})})

\int_{6}^{10} \frac{1}{( x - 6 ) ^{4}} d x

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