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

The answer to (\partial)/(\partial y)(e^{xyz^6}yz^6) is z^6(e^{xyz^6}xyz^6+e^{xyz^6})

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

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

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

Solution steps

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

Treat

x, z

as constants

Take the constant out:

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

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

Apply the Product Rule:

(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}})

Simplify

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

\frac{d}{d x} (7 x + 5 x)

1 - x

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

cos^{2} (1 t)

f (x) = \frac{1}{x - 1}, at x = 5

What is (\partial)/(\partial y)(e^{xyz^6}yz^6) ?
The answer to (\partial)/(\partial y)(e^{xyz^6}yz^6) is z^6(e^{xyz^6}xyz^6+e^{xyz^6})