What is expand 1/3 ((e^{x/3}\H(x))/3+e^{x/3}u'(x)x) ?

The solution to expand 1/3 ((e^{x/3}\H(x))/3+e^{x/3}u'(x)x) is (e^{x/3}\H(x))/9+(e^{x/3}xu'(x))/3

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expand 1/3 ((e^{x/3}\H(x))/3+e^{x/3}u'(x)x)

expand

\frac{1}{3} (\frac{e ^{\frac{x}{3}} H ( x )}{3} + e^{\frac{x}{3}} u^{'} (x) x)

\frac{e ^{\frac{x}{3}} H ( x )}{9} + \frac{e ^{\frac{x}{3}} xu ' ( x )}{3}

Solution steps

\frac{1}{3} (\frac{e ^{\frac{x}{3}} H ( x )}{3} + e^{\frac{x}{3}} u^{'} (x) x)

= \frac{1}{3} (\frac{e ^{\frac{x}{3}} H ( x )}{3} + e^{\frac{x}{3}} x u^{^{'}} (x))

Apply the distributive law:

a (b + c) = ab + a c

a = \frac{1}{3}, b = \frac{e ^{\frac{x}{3}} H ( x )}{3}, c = e^{\frac{x}{3}} u^{'} (x) x

= \frac{1}{3} \cdot \frac{e ^{\frac{x}{3}} H ( x )}{3} + \frac{1}{3} e^{\frac{x}{3}} u^{^{'}} (x) x

= \frac{1}{3} \cdot \frac{e ^{\frac{x}{3}} H ( x )}{3} + \frac{1}{3} e^{\frac{x}{3}} x u^{^{'}} (x)

Simplify

\frac{1}{3} \cdot \frac{e ^{\frac{x}{3}} H ( x )}{3} + \frac{1}{3} e^{\frac{x}{3}} x u^{'} (x) : \frac{e ^{\frac{x}{3}} H ( x )}{9} + \frac{e ^{\frac{x}{3}} x u ^{'} ( x )}{3}

= \frac{e ^{\frac{x}{3}} H ( x )}{9} + \frac{e ^{\frac{x}{3}} x u ^{'} ( x )}{3}

(x^{2} - y^{2}) (e^{- x^{2} - y^{2}})

e^{\frac{x}{3}} (x^{2} - x - 2)

\frac{1}{( x ^{4} + 1 ) ^{2}}

(x - 2) (\cdot + 2)

\frac{8 a ^{2} b - 32 b ^{3}}{x ^{2} + 2 ( a - 2 b ) x - 8 ab}

What is expand 1/3 ((e^{x/3}\H(x))/3+e^{x/3}u'(x)x) ?
The solution to expand 1/3 ((e^{x/3}\H(x))/3+e^{x/3}u'(x)x) is (e^{x/3}\H(x))/9+(e^{x/3}xu'(x))/3