What is the solution for -derivative of y=xe^x ?

The solution for -derivative of y=xe^x is y=-e^xx+e^x+c_{1}

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Find the Derivative - d/dx y=xe^x

- \frac{d}{d x} y = x e^{x}

y = - e^{x} x + e^{x} + c_{1}

Solution steps

- \frac{d y}{d x} = x e^{x}

Isolate

y^{'} (x) : y^{'} (x) = - e^{x} x

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int - e^{x} x d x

\int - e^{x} x d x = - e^{x} x + e^{x} + c_{1}

y = - e^{x} x + e^{x} + c_{1}

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

X

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

f (x) = sin (x)

- \frac{d}{d x} (cos^{2} (x))

What is the solution for -derivative of y=xe^x ?
The solution for -derivative of y=xe^x is y=-e^xx+e^x+c_{1}