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

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

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

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

- e^{x} x - e^{x}

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

Simplify

= - e^{x} x - e^{x}

\frac{d}{d x} ((4 x^{2} - 3 x) e^{5 x})

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

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

n = 0 \sum \infty \frac{2 n - 1}{n !}

\frac{d}{d x} (- \frac{x}{4})

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