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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial x} (4 y^{4}) = 0

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

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

Simplify

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

\int 5 cos^{4} (x) d x

\frac{\partial}{\partial z} (- 3 z)

x \cdot cos (\frac{3 π x}{2}), 0

\int_{0}^{\frac{4}{5}} \frac{1}{16 + 25 x ^{2}} d x

y = sin^{2} (x) \frac{d y}{d x}

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