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

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

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

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

3 y^{2} - 4 x^{3}

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

= 3 y^{2} - 0 - 4 x^{3}

Simplify

= 3 y^{2} - 4 x^{3}

x^{2} y^{^{''}} = y^{^{'}} + (y^{^{'}})^{2}

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

\int \frac{1}{x ^{4} - 16} d x

\int_{0}^{1} 5 e^{5 x} d x

\frac{d}{d x} (\frac{9 x}{x})

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