What is (\partial)/(\partial x)(e^x(x^2-y^2)) ?

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

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(\partial)/(\partial x)(e^x(x^2-y^2))

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

e^{x} (x^{2} - y^{2}) + 2 x e^{x}

Solution steps

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

Treat

y

as a constant

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = e^{x}, g = x^{2} - y^{2}

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

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

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

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

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

\frac{d}{d x} (\frac{3 x ^{3}}{2})

\frac{d}{d x} (sin (x) + x^{2})

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

(y e^{x y} cos (2 x) - 2 e^{x y} sin (2 x) + 2 x) + (x e^{x y} cos (2 x) - 3) y^{^{'}} = 0

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