What is the derivative of f(x)=2x^2e^x(x^2-3x+1) ?

The derivative of f(x)=2x^2e^x(x^2-3x+1) is 2(e^xx^4+e^xx^3-8e^xx^2+2e^xx)

What is the first derivative of f(x)=2x^2e^x(x^2-3x+1) ?

The first derivative of f(x)=2x^2e^x(x^2-3x+1) is 2(e^xx^4+e^xx^3-8e^xx^2+2e^xx)

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derivative of

f (x) = 2 x^{2} e^{x} (x^{2} - 3 x + 1)

2 (e^{x} x^{4} + e^{x} x^{3} - 8 e^{x} x^{2} + 2 e^{x} x)

Solution steps

\frac{d}{d x} (2 x^{2} e^{x} (x^{2} - 3 x + 1))

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d}{d x} (x^{2} e^{x} (x^{2} - 3 x + 1))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{2}, g = e^{x} (x^{2} - 3 x + 1)

= 2 (\frac{d}{d x} (x^{2}) e^{x} (x^{2} - 3 x + 1) + \frac{d}{d x} (e^{x} (x^{2} - 3 x + 1)) x^{2})

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

\frac{d}{d x} (e^{x} (x^{2} - 3 x + 1)) = e^{x} x^{2} - e^{x} x - 2 e^{x}

= 2 (2 x e^{x} (x^{2} - 3 x + 1) + (e^{x} x^{2} - e^{x} x - 2 e^{x}) x^{2})

Expand

2 x e^{x} (x^{2} - 3 x + 1) + (e^{x} x^{2} - e^{x} x - 2 e^{x}) x^{2} : e^{x} x^{4} + e^{x} x^{3} - 8 e^{x} x^{2} + 2 e^{x} x

= 2 (e^{x} x^{4} + e^{x} x^{3} - 8 e^{x} x^{2} + 2 e^{x} x)

\int 2 x \cdot e^{x} d x

\int t^{8} d t

\int_{1}^{4} \frac{x - 7}{x} d x

\frac{\partial}{\partial x} (2 (1 - e^{x})^{2})

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

What is the derivative of f(x)=2x^2e^x(x^2-3x+1) ?
The derivative of f(x)=2x^2e^x(x^2-3x+1) is 2(e^xx^4+e^xx^3-8e^xx^2+2e^xx)
What is the first derivative of f(x)=2x^2e^x(x^2-3x+1) ?
The first derivative of f(x)=2x^2e^x(x^2-3x+1) is 2(e^xx^4+e^xx^3-8e^xx^2+2e^xx)