What is (\partial)/(\partial x)(xe^{xy}-2y+1) ?

The answer to (\partial)/(\partial x)(xe^{xy}-2y+1) is e^{xy}+e^{xy}xy

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(\partial)/(\partial x)(xe^{xy}-2y+1)

\frac{\partial}{\partial x} (x e^{x y} - 2 y + 1)

e^{x y} + e^{x y} x y

Solution steps

\frac{\partial}{\partial x} (x e^{x y} - 2 y + 1)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (x e^{x y}) - \frac{\partial}{\partial x} (2 y) + \frac{\partial}{\partial x} (1)

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

\frac{\partial}{\partial x} (2 y) = 0

\frac{\partial}{\partial x} (1) = 0

= e^{y x} + y e^{y x} x - 0 + 0

Simplify

= e^{x y} + e^{x y} x y

x \to 3 lim (cos (\frac{1}{x - 3}))

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

\frac{d}{d x} (0.005 x^{2} - ln (x^{16}) + 70)

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

- 4 x^{\frac{1}{2}}

What is (\partial)/(\partial x)(xe^{xy}-2y+1) ?
The answer to (\partial)/(\partial x)(xe^{xy}-2y+1) is e^{xy}+e^{xy}xy