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

The answer to (\partial)/(\partial y)((xyz^2)/(xy+1)) is (xz^2)/((xy+1)^2)

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

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

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

Solution steps

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

Treat

x, z

as constants

Take the constant out:

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

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

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

Simplify

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

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

f (x) = 6 x, (6, 6)

\frac{30 s + 102}{s ^{2} + 4 s + 40}

\int_{4}^{\infty} e^{- 4 x} d x

\frac{d}{d x} ((4 x - 2)^{2})

\frac{\partial}{\partial s} (\frac{( s - 3 r )}{t})

What is (\partial)/(\partial y)((xyz^2)/(xy+1)) ?
The answer to (\partial)/(\partial y)((xyz^2)/(xy+1)) is (xz^2)/((xy+1)^2)