What is the derivative of x^2e^x-2xe^x+6e^x ?

The derivative of x^2e^x-2xe^x+6e^x is e^xx^2+4e^x

What is the first derivative of x^2e^x-2xe^x+6e^x ?

The first derivative of x^2e^x-2xe^x+6e^x is e^xx^2+4e^x

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derivative of

x^{2} e^{x} - 2 x e^{x} + 6 e^{x}

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

Solution steps

\frac{d}{d x} (x^{2} e^{x} - 2 x e^{x} + 6 e^{x})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (x^{2} e^{x}) - \frac{d}{d x} (2 x e^{x}) + \frac{d}{d x} (6 e^{x})

\frac{d}{d x} (x^{2} e^{x}) = 2 x e^{x} + e^{x} x^{2}

\frac{d}{d x} (2 x e^{x}) = 2 (e^{x} + e^{x} x)

\frac{d}{d x} (6 e^{x}) = 6 e^{x}

= 2 x e^{x} + e^{x} x^{2} - 2 (e^{x} + e^{x} x) + 6 e^{x}

Simplify

2 x e^{x} + e^{x} x^{2} - 2 (e^{x} + e^{x} x) + 6 e^{x} : e^{x} x^{2} + 4 e^{x}

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

\int_{2}^{2 x} 15 - t d t

y = x^{5} cos (x)

\frac{d}{d x} (a (x + 1) (x - r))

t y^{^{'}} + y = 0

\frac{d x}{d t} = x - 4

What is the derivative of x^2e^x-2xe^x+6e^x ?
The derivative of x^2e^x-2xe^x+6e^x is e^xx^2+4e^x
What is the first derivative of x^2e^x-2xe^x+6e^x ?
The first derivative of x^2e^x-2xe^x+6e^x is e^xx^2+4e^x