What is the derivative of x^2e^x-2xe^x+8e^x ?

The derivative of x^2e^x-2xe^x+8e^x is e^xx^2+6e^x

What is the first derivative of x^2e^x-2xe^x+8e^x ?

The first derivative of x^2e^x-2xe^x+8e^x is e^xx^2+6e^x

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\frac{d}{d x} (x^{2} e^{x} - 2 x e^{x} + 8 e^{x})

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

Solution steps

\frac{d}{d x} (x^{2} e^{x} - 2 x e^{x} + 8 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} (8 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} (8 e^{x}) = 8 e^{x}

= 2 x e^{x} + e^{x} x^{2} - 2 (e^{x} + e^{x} x) + 8 e^{x}

Simplify

2 x e^{x} + e^{x} x^{2} - 2 (e^{x} + e^{x} x) + 8 e^{x} : e^{x} x^{2} + 6 e^{x}

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

\frac{\partial}{\partial x} (1 - x y)

x \to 0 lim (\frac{6 ^{x} - 1}{x})

\frac{d}{d x} (e^{x - 7})

y^{^{''}} - y^{^{'}} = 2 x

f (x) = e - x^{e} + e^{41} - ln (2)

What is the derivative of x^2e^x-2xe^x+8e^x ?
The derivative of x^2e^x-2xe^x+8e^x is e^xx^2+6e^x
What is the first derivative of x^2e^x-2xe^x+8e^x ?
The first derivative of x^2e^x-2xe^x+8e^x is e^xx^2+6e^x