What is (\partial)/(\partial x)(x^6-4x^2y^4+3y^2) ?

The answer to (\partial)/(\partial x)(x^6-4x^2y^4+3y^2) is 6x^5-8xy^4

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(\partial)/(\partial x)(x^6-4x^2y^4+3y^2)

\frac{\partial}{\partial x} (x^{6} - 4 x^{2} y^{4} + 3 y^{2})

6 x^{5} - 8 x y^{4}

Solution steps

\frac{\partial}{\partial x} (x^{6} - 4 x^{2} y^{4} + 3 y^{2})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (x^{6}) - \frac{\partial}{\partial x} (4 x^{2} y^{4}) + \frac{\partial}{\partial x} (3 y^{2})

\frac{\partial}{\partial x} (x^{6}) = 6 x^{5}

\frac{\partial}{\partial x} (4 x^{2} y^{4}) = 8 y^{4} x

\frac{\partial}{\partial x} (3 y^{2}) = 0

= 6 x^{5} - 8 y^{4} x + 0

Simplify

= 6 x^{5} - 8 x y^{4}

\frac{d}{d x} ((- x^{3} + 1) (- 2 x^{2} - 2 x))

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

y^{^{''}} + 6 y^{^{'}} + 5 y = 16 t e^{5 t}

y^{^{''}} + 6 y^{^{'}} - 4 y = 0

ln (1 0^{x})

What is (\partial)/(\partial x)(x^6-4x^2y^4+3y^2) ?
The answer to (\partial)/(\partial x)(x^6-4x^2y^4+3y^2) is 6x^5-8xy^4