What is (\partial)/(\partial x)(1+x*ln(xy-5)) ?

The answer to (\partial)/(\partial x)(1+x*ln(xy-5)) is (xy)/(xy-5)+ln(xy-5)

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(\partial)/(\partial x)(1+x*ln(xy-5))

\frac{\partial}{\partial x} (1 + x \cdot ln (x y - 5))

\frac{x y}{x y - 5} + ln (x y - 5)

Solution steps

\frac{\partial}{\partial x} (1 + x ln (x y - 5))

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} (x ln (x y - 5))

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

\frac{\partial}{\partial x} (x ln (x y - 5)) = ln (y x - 5) + \frac{y x}{y x - 5}

= 0 + ln (y x - 5) + \frac{y x}{y x - 5}

Simplify

= \frac{x y}{x y - 5} + ln (x y - 5)

y^{^{'}} + x^{2} y = x^{2}

3 (cos (x^{2}) - 2 x^{2} sin (x^{2}))

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

y = ln (lo g_{e} (3 x))

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What is (\partial)/(\partial x)(1+x*ln(xy-5)) ?
The answer to (\partial)/(\partial x)(1+x*ln(xy-5)) is (xy)/(xy-5)+ln(xy-5)