What is (\partial)/(\partial x)(x^4y^3+8x^2y) ?

The answer to (\partial)/(\partial x)(x^4y^3+8x^2y) is 4y^3x^3+16yx

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(\partial)/(\partial x)(x^4y^3+8x^2y)

\frac{\partial}{\partial x} (x^{4} y^{3} + 8 x^{2} y)

4 y^{3} x^{3} + 16 y x

Solution steps

\frac{\partial}{\partial x} (x^{4} y^{3} + 8 x^{2} y)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (x^{4} y^{3}) + \frac{\partial}{\partial x} (8 x^{2} y)

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

\frac{\partial}{\partial x} (8 x^{2} y) = 16 y x

= 4 y^{3} x^{3} + 16 y x

\frac{d}{d x} (\frac{x - 5}{5})

81 - y^{2}

y^{^{'}} = 6 - 3 y

x \to 0 lim (\frac{tan ( 12 x )}{x})

x \to \infty lim (3 x + 7)

What is (\partial)/(\partial x)(x^4y^3+8x^2y) ?
The answer to (\partial)/(\partial x)(x^4y^3+8x^2y) is 4y^3x^3+16yx