What is (\partial)/(\partial x)(1+ln(xy)) ?

The answer to (\partial)/(\partial x)(1+ln(xy)) is 1/x

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(\partial)/(\partial x)(1+ln(xy))

\frac{\partial}{\partial x} (1 + ln (x y))

\frac{1}{x}

Solution steps

\frac{\partial}{\partial x} (1 + ln (x y))

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 0 + \frac{1}{x}

Simplify

= \frac{1}{x}

y^{^{''}} - 8 y^{^{'}} + 16 y = t^{- 2} e^{4 t}

x y^{2} y^{^{'}} = y^{3} - 3 x^{3}

\frac{\partial}{\partial y} (\frac{x}{8 + y})

\frac{d}{d x} (lo g_{\frac{1}{x}} (e))

\int \frac{1 + 3 x}{x} d x

What is (\partial)/(\partial x)(1+ln(xy)) ?
The answer to (\partial)/(\partial x)(1+ln(xy)) is 1/x