What is (\partial)/(\partial x)(x^2y^2-y^3+3x^4+5) ?

The answer to (\partial)/(\partial x)(x^2y^2-y^3+3x^4+5) is 12x^3+2xy^2

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(\partial)/(\partial x)(x^2y^2-y^3+3x^4+5)

\frac{\partial}{\partial x} (x^{2} y^{2} - y^{3} + 3 x^{4} + 5)

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

Solution steps

\frac{\partial}{\partial x} (x^{2} y^{2} - y^{3} + 3 x^{4} + 5)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (x^{2} y^{2}) - \frac{\partial}{\partial x} (y^{3}) + \frac{\partial}{\partial x} (3 x^{4}) + \frac{\partial}{\partial x} (5)

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

\frac{\partial}{\partial x} (y^{3}) = 0

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

\frac{\partial}{\partial x} (5) = 0

= 2 y^{2} x - 0 + 12 x^{3} + 0

Simplify

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

\int_{0}^{4} 10 x d x

f (x) = \frac{1}{x ^{2}}

y^{^{''}} + 2 k^{2} y^{^{'}} + k^{4} y = 0

\frac{d}{d x} (\frac{2 x sin ( x )}{2 + cos ( x )})

\int \frac{x ^{5} + 2 x ^{2} - 1}{x ^{4}} d x

What is (\partial)/(\partial x)(x^2y^2-y^3+3x^4+5) ?
The answer to (\partial)/(\partial x)(x^2y^2-y^3+3x^4+5) is 12x^3+2xy^2