What is (\partial)/(\partial x)(((3x^4-2xy))/(y^2)) ?

The answer to (\partial)/(\partial x)(((3x^4-2xy))/(y^2)) is (12x^3-2y)/(y^2)

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(\partial)/(\partial x)(((3x^4-2xy))/(y^2))

\frac{\partial}{\partial x} (\frac{( 3 x ^{4} - 2 x y )}{y ^{2}})

\frac{12 x ^{3} - 2 y}{y ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{( 3 x ^{4} - 2 x y )}{y ^{2}})

Treat

y

as a constant

Take the constant out:

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

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{\partial}{\partial x} (3 x^{4}) = 12 x^{3}

\frac{\partial}{\partial x} (2 x y) = 2 y

= \frac{1}{y ^{2}} (12 x^{3} - 2 y)

Simplify

\frac{1}{y ^{2}} (12 x^{3} - 2 y) : \frac{12 x ^{3} - 2 y}{y ^{2}}

= \frac{12 x ^{3} - 2 y}{y ^{2}}

\int \frac{2 x}{x ^{4} + 4} d x

\frac{d y}{d x} = \frac{6 x}{y}

x^{2} + 1

\int \frac{3 x}{x ^{2} + 6 x + 9} d x

\frac{d}{d x} (in (2 x))

What is (\partial)/(\partial x)(((3x^4-2xy))/(y^2)) ?
The answer to (\partial)/(\partial x)(((3x^4-2xy))/(y^2)) is (12x^3-2y)/(y^2)