What is (\partial)/(\partial v)({u}(v,w)+ve^w) ?

The answer to (\partial)/(\partial v)({u}(v,w)+ve^w) is (\partial)/(\partial v)({u}(v,w))+e^w

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(\partial)/(\partial v)({u}(v,w)+ve^w)

\frac{\partial}{\partial v} (u (v, w) + v e^{w})

\frac{\partial}{\partial v} (u (v, w)) + e^{w}

Solution steps

\frac{\partial}{\partial v} (u (v, w) + v e^{w})

Treat

w

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial v} (u (v, w)) + \frac{\partial}{\partial v} (v e^{w})

\frac{\partial}{\partial v} (v e^{w}) = e^{w}

= \frac{\partial}{\partial v} (u (v, w)) + e^{w}

n = 2 \sum \infty \frac{8}{n ln ( n )}

\frac{d y}{d x} = \frac{e ^{2 x - y}}{e ^{x + y}}

\int \frac{1}{1 - 2 x} - cos (\frac{x}{3}) d x

\frac{\partial}{\partial x} (yz e^{x})

What is (\partial)/(\partial v)({u}(v,w)+ve^w) ?
The answer to (\partial)/(\partial v)({u}(v,w)+ve^w) is (\partial)/(\partial v)({u}(v,w))+e^w