What is the derivative of f(x)=x^2e^x-2xe^x+2e^x ?

The derivative of f(x)=x^2e^x-2xe^x+2e^x is e^xx^2

What is the first derivative of f(x)=x^2e^x-2xe^x+2e^x ?

The first derivative of f(x)=x^2e^x-2xe^x+2e^x is e^xx^2

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derivative of

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

e^{x} x^{2}

Solution steps

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

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

Simplify

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

= e^{x} x^{2}

\int tan^{4} (5 t) d t

\int \frac{1 + 3 x}{( 1 - x ) ( 3 x - 5 )} d x

\int e^{7 t} d t

y = 5 + 5 x^{2} - 2 x^{3}, (2, 9)

\frac{d y}{d x} = y - y^{3}

What is the derivative of f(x)=x^2e^x-2xe^x+2e^x ?
The derivative of f(x)=x^2e^x-2xe^x+2e^x is e^xx^2
What is the first derivative of f(x)=x^2e^x-2xe^x+2e^x ?
The first derivative of f(x)=x^2e^x-2xe^x+2e^x is e^xx^2