What is the derivative of 9x^2e^x ?

The derivative of 9x^2e^x is 9(2xe^x+e^xx^2)

What is the first derivative of 9x^2e^x ?

The first derivative of 9x^2e^x is 9(2xe^x+e^xx^2)

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Product Rule:

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

f = x^{2}, g = e^{x}

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

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

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

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

\frac{d}{d x} (x sinh (x) - cosh (x))

\int e^{2} - 2 cos (x) d x

n = 1 \sum \infty \frac{1}{n ^{\frac{1}{2}}}

y = 12 cos (2) (x)

\int \frac{3 x}{x ^{2} + 5} d x

What is the derivative of 9x^2e^x ?
The derivative of 9x^2e^x is 9(2xe^x+e^xx^2)
What is the first derivative of 9x^2e^x ?
The first derivative of 9x^2e^x is 9(2xe^x+e^xx^2)