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

The answer to (\partial)/(\partial y)(x^4+y^4+3x^2y^3) is 4y^3+9x^2y^2

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

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

4 y^{3} + 9 x^{2} y^{2}

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

\frac{\partial}{\partial y} (3 x^{2} y^{3}) = 9 x^{2} y^{2}

= 0 + 4 y^{3} + 9 x^{2} y^{2}

Simplify

= 4 y^{3} + 9 x^{2} y^{2}

\frac{\partial}{\partial x} (x e^{- x^{2}})

\frac{1}{s + 1 ( - 2 s + 10 )}

n = 1 \sum \infty \frac{( - 5 ) ^{n}}{3 ^{2 n + 1}}

\frac{d}{d x} (32 cos (x))

n = 1 \sum \infty \frac{x ^{n}}{1 6 ^{n}}

What is (\partial)/(\partial y)(x^4+y^4+3x^2y^3) ?
The answer to (\partial)/(\partial y)(x^4+y^4+3x^2y^3) is 4y^3+9x^2y^2