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

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

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

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

- 4 x^{3} + 4 x

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

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

= 0 - 4 x^{3} + 4 x - 0

Simplify

= - 4 x^{3} + 4 x

\int_{- \frac{π}{6}}^{0} 2 sec^{3} (x) d x

\frac{\partial}{\partial y} (x^{a} y^{b} z^{c})

f (x) = \frac{3}{x ^{2} + 3}

\frac{e ^{x} - e ^{- x}}{2}

\frac{d}{d x} (ln (e^{3 x^{2}} \cdot 4^{5 x}))

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