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

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

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

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

12 x^{2} y^{3}

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the Power Rule:

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

= 3 x^{2} \cdot 4 y^{4 - 1}

Simplify

= 12 x^{2} y^{3}

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

x \to 0 lim (sin (- \frac{1}{x}))

f (x) = x^{3} - 3 x^{2} + 2

\frac{d y}{d t} = \frac{( t e ^{t} )}{y 5 + y ^{2}}

\frac{\partial}{\partial x} (c x)

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