What is the derivative of y=x^2e^x-2xe^x+7e^x ?

The derivative of y=x^2e^x-2xe^x+7e^x is e^xx^2+5e^x

What is the first derivative of y=x^2e^x-2xe^x+7e^x ?

The first derivative of y=x^2e^x-2xe^x+7e^x is e^xx^2+5e^x

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derivative of

y = x^{2} e^{x} - 2 x e^{x} + 7 e^{x}

e^{x} x^{2} + 5 e^{x}

Solution steps

\frac{d}{d x} (x^{2} e^{x} - 2 x e^{x} + 7 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} (7 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} (7 e^{x}) = 7 e^{x}

= 2 x e^{x} + e^{x} x^{2} - 2 (e^{x} + e^{x} x) + 7 e^{x}

Simplify

2 x e^{x} + e^{x} x^{2} - 2 (e^{x} + e^{x} x) + 7 e^{x} : e^{x} x^{2} + 5 e^{x}

= e^{x} x^{2} + 5 e^{x}

x^{2} y^{^{''}} + 7 x y^{^{'}} + 8 y = \frac{6}{x}

f (x) = 8 x^{4}

\int \frac{( 6 + 6 x )}{1 + x ^{2}} d x

\int x^{2} cos (x) + 2 sin (x) d x

\int_{- \infty}^{0} x e^{7 x} d x

What is the derivative of y=x^2e^x-2xe^x+7e^x ?
The derivative of y=x^2e^x-2xe^x+7e^x is e^xx^2+5e^x
What is the first derivative of y=x^2e^x-2xe^x+7e^x ?
The first derivative of y=x^2e^x-2xe^x+7e^x is e^xx^2+5e^x