What is the derivative of 3(2xe^x+e^xx^2) ?

The derivative of 3(2xe^x+e^xx^2) is 3(e^xx^2+4e^xx+2e^x)

What is the first derivative of 3(2xe^x+e^xx^2) ?

The first derivative of 3(2xe^x+e^xx^2) is 3(e^xx^2+4e^xx+2e^x)

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

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

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

Expand

2 (e^{x} + e^{x} x) + e^{x} x^{2} + 2 x e^{x} : e^{x} x^{2} + 4 e^{x} x + 2 e^{x}

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

y^{^{'}} = \frac{y ^{2}}{t ^{3}}

ln (x e^{- 6 x})

\frac{sec ( θ )}{1 + sec ( θ )}

\int \frac{7 x ^{2} + 3 x + 4}{x ^{3} - x} d x

\frac{2}{3 x + 1}, 2

What is the derivative of 3(2xe^x+e^xx^2) ?
The derivative of 3(2xe^x+e^xx^2) is 3(e^xx^2+4e^xx+2e^x)
What is the first derivative of 3(2xe^x+e^xx^2) ?
The first derivative of 3(2xe^x+e^xx^2) is 3(e^xx^2+4e^xx+2e^x)