What is (\partial)/(\partial x)(2xe^x-y+6x^2) ?

The answer to (\partial)/(\partial x)(2xe^x-y+6x^2) is 12x+2(e^xx+e^x)

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(\partial)/(\partial x)(2xe^x-y+6x^2)

\frac{\partial}{\partial x} (2 x e^{x} - y + 6 x^{2})

12 x + 2 (e^{x} x + e^{x})

Solution steps

\frac{\partial}{\partial x} (2 x e^{x} - y + 6 x^{2})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial x} (y) = 0

\frac{\partial}{\partial x} (6 x^{2}) = 12 x

= 2 (e^{x} + e^{x} x) - 0 + 12 x

Simplify

= 12 x + 2 (e^{x} x + e^{x})

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

\frac{d}{d x} (x i)

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

3 (x + 1), 4 (x + 1), 0, 3

\int e^{- x} \cdot e^{x} d x

What is (\partial)/(\partial x)(2xe^x-y+6x^2) ?
The answer to (\partial)/(\partial x)(2xe^x-y+6x^2) is 12x+2(e^xx+e^x)