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

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

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

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

12 y^{3}

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial y} (4 x^{2}) = 0

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

= 0 + 12 y^{3}

Simplify

= 12 y^{3}

x \to 0 lim (\frac{tan ( x ) - x}{x})

\int cot^{2} (3 x) d x

\int (1 + x)^{3} d x

\int_{0}^{5} \frac{5 - x}{5} d x

\frac{d}{d x} (\frac{x ^{3} - 1}{x ^{3} + 1})

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