What is (\partial)/(\partial x)(7+ln(xy-9)) ?

The answer to (\partial)/(\partial x)(7+ln(xy-9)) is y/(xy-9)

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(\partial)/(\partial x)(7+ln(xy-9))

\frac{\partial}{\partial x} (7 + ln (x y - 9))

\frac{y}{x y - 9}

Solution steps

\frac{\partial}{\partial x} (7 + ln (x y - 9))

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (7) + \frac{\partial}{\partial x} (ln (x y - 9))

\frac{\partial}{\partial x} (7) = 0

\frac{\partial}{\partial x} (ln (x y - 9)) = \frac{y}{y x - 9}

= 0 + \frac{y}{y x - 9}

Simplify

= \frac{y}{x y - 9}

\frac{d}{d x} (e^{a x^{9}})

\frac{d}{d y} (e^{x} cos (y))

\frac{\partial}{\partial x} (\frac{26}{( 2 + x ^{2} + y ^{2} )})

\int e^{x} \cdot 7 x d x

(u^{2} + u) d x + u^{2} (x d u + u d x) = 0

What is (\partial)/(\partial x)(7+ln(xy-9)) ?
The answer to (\partial)/(\partial x)(7+ln(xy-9)) is y/(xy-9)