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

The answer to (\partial)/(\partial x)(xe^y-ln(x)) is e^y-1/x

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

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

e^{y} - \frac{1}{x}

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= e^{y} - \frac{1}{x}

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

\frac{d}{d x} (\frac{4 x ^{2} - 3}{x - 1})

\frac{d}{d t} ((sin (t) + cos (t))^{2})

y = 5 x^{2} + \frac{6}{x}

y = \frac{8 t}{8 + t}

What is (\partial)/(\partial x)(xe^y-ln(x)) ?
The answer to (\partial)/(\partial x)(xe^y-ln(x)) is e^y-1/x