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

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

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

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

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

Solution steps

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

Treat

v

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

\frac{x ^{3}}{3} + \frac{1}{4 x}

f (x) = \frac{2}{5 x ^{2}}

\frac{\partial}{\partial x} (\frac{( 1 + x )}{1 + y} z^{2})

y^{^{''}} - 2 y^{^{'}} - 35 y = 0

\int \frac{x ^{2}}{16 + x ^{6}} d x

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