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

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

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

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

- 4 x^{3} + 4 x

Solution steps

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

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

(cos (9 x))^{x}

48 y^{^{''}} + 34 y^{^{'}} - 5 y = 0

\frac{d}{d x} ((4 x - 3)^{2})

x^{(\frac{4}{3})}, 2 x^{(\frac{1}{3})}, 0, 2

(e^{y} + 1)^{2} e^{- y} d x + (e^{x} + 1)^{5} e^{- x} d y = 0

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