What is the derivative of y=x^2e^x-2xe^x+9e^x ?

The derivative of y=x^2e^x-2xe^x+9e^x is e^xx^2+7e^x

What is the first derivative of y=x^2e^x-2xe^x+9e^x ?

The first derivative of y=x^2e^x-2xe^x+9e^x is e^xx^2+7e^x

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derivative of

y = x^{2} e^{x} - 2 x e^{x} + 9 e^{x}

e^{x} x^{2} + 7 e^{x}

Solution steps

\frac{d}{d x} (x^{2} e^{x} - 2 x e^{x} + 9 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} (9 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} (9 e^{x}) = 9 e^{x}

= 2 x e^{x} + e^{x} x^{2} - 2 (e^{x} + e^{x} x) + 9 e^{x}

Simplify

2 x e^{x} + e^{x} x^{2} - 2 (e^{x} + e^{x} x) + 9 e^{x} : e^{x} x^{2} + 7 e^{x}

= e^{x} x^{2} + 7 e^{x}

\frac{d y}{d x} - 2 y = - 1

y^{^{'}} = 0.2 x

\int_{1}^{e^{3}} x^{5} ln (x) d x

\frac{d}{d x} (8 x^{6})

x \to \infty lim (e^{2 x})

What is the derivative of y=x^2e^x-2xe^x+9e^x ?
The derivative of y=x^2e^x-2xe^x+9e^x is e^xx^2+7e^x
What is the first derivative of y=x^2e^x-2xe^x+9e^x ?
The first derivative of y=x^2e^x-2xe^x+9e^x is e^xx^2+7e^x