What is (\partial)/(\partial y)(e^{xyz^4}yz^4) ?

The answer to (\partial)/(\partial y)(e^{xyz^4}yz^4) is z^4(e^{xyz^4}xyz^4+e^{xyz^4})

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(\partial)/(\partial y)(e^{xyz^4}yz^4)

\frac{\partial}{\partial y} (e^{x y z^{4}} y z^{4})

z^{4} (e^{x y z^{4}} x y z^{4} + e^{x y z^{4}})

Solution steps

\frac{\partial}{\partial y} (e^{x y z^{4}} y z^{4})

Treat

x, z

as constants

Take the constant out:

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

= z^{4} \frac{\partial}{\partial y} (e^{x y z^{4}} y)

Apply the Product Rule:

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

f = e^{x y z^{4}}, g = y

= z^{4} (\frac{\partial}{\partial y} (e^{x y z^{4}}) y + \frac{\partial}{\partial y} (y) e^{x y z^{4}})

\frac{\partial}{\partial y} (e^{x y z^{4}}) = e^{x y z^{4}} x z^{4}

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

= z^{4} (e^{x y z^{4}} x z^{4} y + 1 \cdot e^{x y z^{4}})

Simplify

= z^{4} (e^{x y z^{4}} x y z^{4} + e^{x y z^{4}})

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

f (x) = x^{3} - 4 x, (2, 0)

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

x \to 1^{- 1} lim (- 1.6 x - 0.6)

\frac{d}{d x} (cot (3 x^{2} + 5))

What is (\partial)/(\partial y)(e^{xyz^4}yz^4) ?
The answer to (\partial)/(\partial y)(e^{xyz^4}yz^4) is z^4(e^{xyz^4}xyz^4+e^{xyz^4})