What is the derivative of e^xx^4+8e^xx^3+12e^xx^2 ?

The derivative of e^xx^4+8e^xx^3+12e^xx^2 is e^xx^4+12e^xx^3+36e^xx^2+24e^xx

What is the first derivative of e^xx^4+8e^xx^3+12e^xx^2 ?

The first derivative of e^xx^4+8e^xx^3+12e^xx^2 is e^xx^4+12e^xx^3+36e^xx^2+24e^xx

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derivative of

e^{x} x^{4} + 8 e^{x} x^{3} + 12 e^{x} x^{2}

e^{x} x^{4} + 12 e^{x} x^{3} + 36 e^{x} x^{2} + 24 e^{x} x

Solution steps

\frac{d}{d x} (e^{x} x^{4} + 8 e^{x} x^{3} + 12 e^{x} x^{2})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (e^{x} x^{4}) + \frac{d}{d x} (8 e^{x} x^{3}) + \frac{d}{d x} (12 e^{x} x^{2})

\frac{d}{d x} (e^{x} x^{4}) = e^{x} x^{4} + 4 x^{3} e^{x}

\frac{d}{d x} (8 e^{x} x^{3}) = 8 (e^{x} x^{3} + 3 x^{2} e^{x})

\frac{d}{d x} (12 e^{x} x^{2}) = 12 (e^{x} x^{2} + 2 x e^{x})

= e^{x} x^{4} + 4 x^{3} e^{x} + 8 (e^{x} x^{3} + 3 x^{2} e^{x}) + 12 (e^{x} x^{2} + 2 x e^{x})

Simplify

e^{x} x^{4} + 4 x^{3} e^{x} + 8 (e^{x} x^{3} + 3 x^{2} e^{x}) + 12 (e^{x} x^{2} + 2 x e^{x}) : e^{x} x^{4} + 12 e^{x} x^{3} + 36 e^{x} x^{2} + 24 e^{x} x

= e^{x} x^{4} + 12 e^{x} x^{3} + 36 e^{x} x^{2} + 24 e^{x} x

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

\int_{0}^{1} (x^{2} + 1) e^{- x} d x

n = 0 \sum \infty \frac{2}{5 ^{n}}

\frac{d}{d x} (csc (2 x))

\frac{d y}{d x} = - \frac{x}{4 y}

What is the derivative of e^xx^4+8e^xx^3+12e^xx^2 ?
The derivative of e^xx^4+8e^xx^3+12e^xx^2 is e^xx^4+12e^xx^3+36e^xx^2+24e^xx
What is the first derivative of e^xx^4+8e^xx^3+12e^xx^2 ?
The first derivative of e^xx^4+8e^xx^3+12e^xx^2 is e^xx^4+12e^xx^3+36e^xx^2+24e^xx