What is (\partial)/(\partial y)(xy(1/(y^3)+2)) ?

The answer to (\partial)/(\partial y)(xy(1/(y^3)+2)) is x(-2/(y^3)+2)

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(\partial)/(\partial y)(xy(1/(y^3)+2))

\frac{\partial}{\partial y} (x y (\frac{1}{y ^{3}} + 2))

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

Solution steps

\frac{\partial}{\partial y} (x y (\frac{1}{y ^{3}} + 2))

Treat

x

as a constant

Take the constant out:

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

= x \frac{\partial}{\partial y} (y (\frac{1}{y ^{3}} + 2))

Apply the Product Rule:

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

f = y, g = \frac{1}{y ^{3}} + 2

= x (\frac{\partial}{\partial y} (y) (\frac{1}{y ^{3}} + 2) + \frac{\partial}{\partial y} (\frac{1}{y ^{3}} + 2) y)

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

\frac{\partial}{\partial y} (\frac{1}{y ^{3}} + 2) = - \frac{3}{y ^{4}}

= x (1 \cdot (\frac{1}{y ^{3}} + 2) + (- \frac{3}{y ^{4}}) y)

Simplify

x (1 \cdot (\frac{1}{y ^{3}} + 2) + (- \frac{3}{y ^{4}}) y) : x (- \frac{2}{y ^{3}} + 2)

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

\int \frac{1}{4} e^{4 x + 1} d x

sin^{2} (π x)

\frac{d}{d x} (\frac{8}{x} - \frac{4}{x})

\frac{\partial}{\partial y} (yz - ln (x + z))

x \to 2 lim (3 x^{2} + 1)

What is (\partial)/(\partial y)(xy(1/(y^3)+2)) ?
The answer to (\partial)/(\partial y)(xy(1/(y^3)+2)) is x(-2/(y^3)+2)