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

The answer to (\partial)/(\partial x)(3x^2+xy^3-2y^2+3) is 6x+y^3

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

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

6 x + y^{3}

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (3 x^{2}) = 6 x

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

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

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

= 6 x + y^{3} - 0 + 0

Simplify

= 6 x + y^{3}

y = tan (x), at x = - \frac{7 π}{4}

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What is (\partial)/(\partial x)(3x^2+xy^3-2y^2+3) ?
The answer to (\partial)/(\partial x)(3x^2+xy^3-2y^2+3) is 6x+y^3