What is (\partial)/(\partial y)(e^{xyz^3}yz^3) ?

The answer to (\partial)/(\partial y)(e^{xyz^3}yz^3) is z^3(e^{xyz^3}xyz^3+e^{xyz^3})

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(\partial)/(\partial y)(e^{xyz^3}yz^3)

\frac{\partial}{\partial y} (e^{x y z^{3}} y z^{3})

z^{3} (e^{x y z^{3}} x y z^{3} + e^{x y z^{3}})

Solution steps

\frac{\partial}{\partial y} (e^{x y z^{3}} y z^{3})

Treat

x, z

as constants

Take the constant out:

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

= z^{3} \frac{\partial}{\partial y} (e^{x y z^{3}} y)

Apply the Product Rule:

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

f = e^{x y z^{3}}, g = y

= z^{3} (\frac{\partial}{\partial y} (e^{x y z^{3}}) y + \frac{\partial}{\partial y} (y) e^{x y z^{3}})

\frac{\partial}{\partial y} (e^{x y z^{3}}) = e^{x y z^{3}} x z^{3}

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

= z^{3} (e^{x y z^{3}} x z^{3} y + 1 \cdot e^{x y z^{3}})

Simplify

= z^{3} (e^{x y z^{3}} x y z^{3} + e^{x y z^{3}})

y = - \frac{1}{4} x + 5

f (x) = 3 cos (x), at x = \frac{π}{4}

\int 4 cos^{3} (4 x) \cdot sin (4 x) d x

\frac{( x ^{2} + 1 ) d y}{d x} + 9 x (y - 1) = 0

\frac{d}{d x} (tan^{2} (x) - 1)

What is (\partial)/(\partial y)(e^{xyz^3}yz^3) ?
The answer to (\partial)/(\partial y)(e^{xyz^3}yz^3) is z^3(e^{xyz^3}xyz^3+e^{xyz^3})