What is the (2xe^{x+3}+e^{x+3}x^2)^' ?

The (2xe^{x+3}+e^{x+3}x^2)^' is e^{x+3}x^2+4e^{x+3}x+2e^{x+3}

What is the first (2xe^{x+3}+e^{x+3}x^2)^' ?

The first (2xe^{x+3}+e^{x+3}x^2)^' is e^{x+3}x^2+4e^{x+3}x+2e^{x+3}

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(2 x e^{x + 3} + e^{x + 3} x^{2})^{^{'}}

e^{x + 3} x^{2} + 4 e^{x + 3} x + 2 e^{x + 3}

Solution steps

(2 x e^{x + 3} + e^{x + 3} x^{2})^{'}

Apply the Sum/Difference Rule:

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

= (2 x e^{x + 3})^{^{'}} + (e^{x + 3} x^{2})^{^{'}}

(2 x e^{x + 3})^{'} = 2 (e^{x + 3} + e^{x + 3} x)

(e^{x + 3} x^{2})^{'} = e^{x + 3} x^{2} + 2 x e^{x + 3}

= 2 (e^{x + 3} + e^{x + 3} x) + e^{x + 3} x^{2} + 2 x e^{x + 3}

Simplify

2 (e^{x + 3} + e^{x + 3} x) + e^{x + 3} x^{2} + 2 x e^{x + 3} : e^{x + 3} x^{2} + 4 e^{x + 3} x + 2 e^{x + 3}

= e^{x + 3} x^{2} + 4 e^{x + 3} x + 2 e^{x + 3}

\frac{d ^{4}}{d x ^{4}} (7 e^{x^{2}})

y^{^{'}} + 5 x^{- 1} y = x^{6}

x \to 3 lim (\frac{3}{x - 3})

ln (7 - x^{2})

n = 0 \sum \infty (\frac{1}{6})^{n + 1}

What is the (2xe^{x+3}+e^{x+3}x^2)^' ?
The (2xe^{x+3}+e^{x+3}x^2)^' is e^{x+3}x^2+4e^{x+3}x+2e^{x+3}
What is the first (2xe^{x+3}+e^{x+3}x^2)^' ?
The first (2xe^{x+3}+e^{x+3}x^2)^' is e^{x+3}x^2+4e^{x+3}x+2e^{x+3}