What is the derivative of y=x^2e^x-2xe^x+6e^x ?

The derivative of y=x^2e^x-2xe^x+6e^x is e^xx^2+4e^x

What is the first derivative of y=x^2e^x-2xe^x+6e^x ?

The first derivative of y=x^2e^x-2xe^x+6e^x is e^xx^2+4e^x

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derivative of

y = x^{2} e^{x} - 2 x e^{x} + 6 e^{x}

e^{x} x^{2} + 4 e^{x}

Solution steps

\frac{d}{d x} (x^{2} e^{x} - 2 x e^{x} + 6 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} (6 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} (6 e^{x}) = 6 e^{x}

= 2 x e^{x} + e^{x} x^{2} - 2 (e^{x} + e^{x} x) + 6 e^{x}

Simplify

2 x e^{x} + e^{x} x^{2} - 2 (e^{x} + e^{x} x) + 6 e^{x} : e^{x} x^{2} + 4 e^{x}

= e^{x} x^{2} + 4 e^{x}

f (x) = 16 x, at x = 16

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

\int \frac{1}{1 0 ^{5} x - x ^{2}} d x

\int 4 x^{\frac{1}{3}} + 2 x^{- \frac{1}{3}} d x

y^{^{'}} = y (x y^{6} + 3)

What is the derivative of y=x^2e^x-2xe^x+6e^x ?
The derivative of y=x^2e^x-2xe^x+6e^x is e^xx^2+4e^x
What is the first derivative of y=x^2e^x-2xe^x+6e^x ?
The first derivative of y=x^2e^x-2xe^x+6e^x is e^xx^2+4e^x