What is (\partial)/(\partial x)((x-4)ln(xy)) ?

The answer to (\partial)/(\partial x)((x-4)ln(xy)) is ln(xy)+(x-4)/x

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(\partial)/(\partial x)((x-4)ln(xy))

\frac{\partial}{\partial x} ((x - 4) ln (x y))

ln (x y) + \frac{x - 4}{x}

Solution steps

\frac{\partial}{\partial x} ((x - 4) ln (x y))

Treat

y

as a constant

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x - 4, g = ln (x y)

= \frac{\partial}{\partial x} (x - 4) ln (x y) + \frac{\partial}{\partial x} (ln (x y)) (x - 4)

\frac{\partial}{\partial x} (x - 4) = 1

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

= 1 \cdot ln (x y) + \frac{1}{x} (x - 4)

Simplify

1 \cdot ln (x y) + \frac{1}{x} (x - 4) : ln (x y) + \frac{x - 4}{x}

= ln (x y) + \frac{x - 4}{x}

\int - x cos (x) + x d x

\int_{0}^{l} cos (\frac{π y}{2 l}) d y

g (t) = t - \frac{4}{t ^{3}}

f (t) = 2 e^{2 t}

\frac{d}{d x} (\frac{x + 1}{2})

What is (\partial)/(\partial x)((x-4)ln(xy)) ?
The answer to (\partial)/(\partial x)((x-4)ln(xy)) is ln(xy)+(x-4)/x