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Beliebt Rechnen >

(\partial)/(\partial y)(e^{xyz^6}yz^6)

Lösung

\frac{\partial}{\partial y} (e^{x y z^{6}} y z^{6})

Lösung

z^{6} (e^{x y z^{6}} x y z^{6} + e^{x y z^{6}})

Schritte zur Lösung

\frac{\partial}{\partial y} (e^{x y z^{6}} y z^{6})

Behandle

x, z

als Konstanten

Entferne die Konstante:

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

= z^{6} \frac{\partial}{\partial y} (e^{x y z^{6}} y)

Wende die Produktregel an:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = e^{x y z^{6}}, g = y

= z^{6} (\frac{\partial}{\partial y} (e^{x y z^{6}}) y + \frac{\partial}{\partial y} (y) e^{x y z^{6}})

\frac{\partial}{\partial y} (e^{x y z^{6}}) = e^{x y z^{6}} x z^{6}

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

= z^{6} (e^{x y z^{6}} x z^{6} y + 1 \cdot e^{x y z^{6}})

Vereinfache

= z^{6} (e^{x y z^{6}} x y z^{6} + e^{x y z^{6}})

Beliebte Beispiele

maclaurin sqrt(1-x)

ma c l a u r in 1 - x

y^{''}+8y^'+16y=1521e^{2t}cos(9t)

y^{^{''}} + 8 y^{^{'}} + 16 y = 1521 e^{2 t} cos (9 t)

laplacetransform cos^2(1t)

l a pl a ce t r an s f or m cos^{2} (1 t)

tangent f(x)= 1/(x-1),\at x=5

t an g e n t f (x) = \frac{1}{x - 1}, at x = 5

derivative of (ln(x)/(x^3))

\frac{d}{d x} (\frac{ln ( x )}{x ^{3}})