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

The answer to (\partial)/(\partial x)(3yuxuv^3-2) is 3yu^2v^3

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(\partial)/(\partial x)(3yuxuv^3-2)

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

3 y u^{2} v^{3}

Solution steps

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

Simplify

3 y uxu v^{3} - 2 : 3 y u^{2} x v^{3} - 2

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

Treat

y, u, v

as constants

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial x} (2) = 0

= 3 y u^{2} v^{3} - 0

Simplify

= 3 y u^{2} v^{3}

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

n = 8 \sum \infty \frac{1}{n} sin (\frac{1}{n})

y^{^{'}} = \frac{t y}{t + 1}, y (4) = 1

(x + y) d x + (3 x + 3 y - 4) d y = 0

What is (\partial)/(\partial x)(3yuxuv^3-2) ?
The answer to (\partial)/(\partial x)(3yuxuv^3-2) is 3yu^2v^3