What is (\partial)/(\partial u)((u+2v)(u-2v)+(u-2v)(uv)+(u+2v)(uv)) ?

The answer to (\partial)/(\partial u)((u+2v)(u-2v)+(u-2v)(uv)+(u+2v)(uv)) is 4uv+2u

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(\partial)/(\partial u)((u+2v)(u-2v)+(u-2v)(uv)+(u+2v)(uv))

\frac{\partial}{\partial u} ((u + 2 v) (u - 2 v) + (u - 2 v) (uv) + (u + 2 v) (uv))

4 uv + 2 u

Solution steps

\frac{\partial}{\partial u} ((u + 2 v) (u - 2 v) + (u - 2 v) (uv) + (u + 2 v) (uv))

Simplify

(u + 2 v) (u - 2 v) + (u - 2 v) (uv) + (u + 2 v) (uv) : 2 u^{2} v + u^{2} - 4 v^{2}

= \frac{\partial}{\partial u} (2 u^{2} v + u^{2} - 4 v^{2})

Treat

v

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial u} (2 u^{2} v) + \frac{\partial}{\partial u} (u^{2}) - \frac{\partial}{\partial u} (4 v^{2})

\frac{\partial}{\partial u} (2 u^{2} v) = 4 vu

\frac{\partial}{\partial u} (u^{2}) = 2 u

\frac{\partial}{\partial u} (4 v^{2}) = 0

= 4 vu + 2 u - 0

Simplify

= 4 uv + 2 u

\int x (x + 5) d x

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

\frac{10}{s} (\frac{1}{10 s + 1})

x^{2} + e^{x}

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

What is (\partial)/(\partial u)((u+2v)(u-2v)+(u-2v)(uv)+(u+2v)(uv)) ?
The answer to (\partial)/(\partial u)((u+2v)(u-2v)+(u-2v)(uv)+(u+2v)(uv)) is 4uv+2u