What is (\partial}{\partial y}(\frac{3x^2)/y) ?

The answer to (\partial}{\partial y}(\frac{3x^2)/y) is -(3x^2)/(y^2)

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

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

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

Simplify

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

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

x \to 5 + lim (\frac{1}{( x - 5 ) ^{2}})

f (x) = \frac{1 + x}{1 - x}

x \to 1 lim (e^{4 x^{2} - 4 x})

\int x^{2} 2 x + 3 d x

f (x) = 3 arccos (\frac{x}{2})

What is (\partial}{\partial y}(\frac{3x^2)/y) ?
The answer to (\partial}{\partial y}(\frac{3x^2)/y) is -(3x^2)/(y^2)