What is the derivative of axe^x+ae^x ?

The derivative of axe^x+ae^x is ae^xx+2ae^x

What is the first derivative of axe^x+ae^x ?

The first derivative of axe^x+ae^x is ae^xx+2ae^x

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\frac{d}{d x} (a x e^{x} + a e^{x})

a e^{x} x + 2 a e^{x}

Solution steps

\frac{d}{d x} (a x e^{x} + a e^{x})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (a x e^{x}) + \frac{d}{d x} (a e^{x})

\frac{d}{d x} (a x e^{x}) = a (e^{x} + e^{x} x)

\frac{d}{d x} (a e^{x}) = a e^{x}

= a (e^{x} + e^{x} x) + a e^{x}

Simplify

a (e^{x} + e^{x} x) + a e^{x} : a e^{x} x + 2 a e^{x}

= a e^{x} x + 2 a e^{x}

\frac{d}{d x} (- 3 x^{4} + 4 x^{3} + 2)

\int 3 e^{- s t} d t

\int - 3 cos (5 x) d x

f (x) = 16 x^{\frac{1}{8}}

y^{^{''}} - 4 y^{^{'}} + 4 y = (12 x^{2} - 6 x) e^{2 x}

What is the derivative of axe^x+ae^x ?
The derivative of axe^x+ae^x is ae^xx+2ae^x
What is the first derivative of axe^x+ae^x ?
The first derivative of axe^x+ae^x is ae^xx+2ae^x