What is (\partial)/(\partial y)(2ln(3x)-4ln(2xy)) ?

The answer to (\partial)/(\partial y)(2ln(3x)-4ln(2xy)) is -4/y

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(\partial)/(\partial y)(2ln(3x)-4ln(2xy))

\frac{\partial}{\partial y} (2 ln (3 x) - 4 ln (2 x y))

- \frac{4}{y}

Solution steps

\frac{\partial}{\partial y} (2 ln (3 x) - 4 ln (2 x y))

Simplify

2 ln (3 x) - 4 ln (2 x y) : ln (\frac{9}{16 x ^{2} y ^{4}})

= \frac{\partial}{\partial y} (ln (\frac{9}{16 x ^{2} y ^{4}}))

Treat

x

as a constant

Apply the chain rule:

\frac{1}{\frac{9}{16 x ^{2} y ^{4}}} \frac{\partial}{\partial y} (\frac{9}{16 x ^{2} y ^{4}})

= \frac{1}{\frac{9}{16 x ^{2} y ^{4}}} \frac{\partial}{\partial y} (\frac{9}{16 x ^{2} y ^{4}})

\frac{\partial}{\partial y} (\frac{9}{16 x ^{2} y ^{4}}) = - \frac{9}{4 x ^{2} y ^{5}}

= \frac{1}{\frac{9}{16 x ^{2} y ^{4}}} (- \frac{9}{4 x ^{2} y ^{5}})

Simplify

\frac{1}{\frac{9}{16 x ^{2} y ^{4}}} (- \frac{9}{4 x ^{2} y ^{5}}) : - \frac{4}{y}

= - \frac{4}{y}

\frac{3}{x ^{3}}

\int 15 x^{4} d x

\frac{3}{s ^{2} - 4}

\frac{d}{d x} (x^{3 l n (x)})

\int - \frac{2}{3} y^{- \frac{4}{3}} d y

What is (\partial)/(\partial y)(2ln(3x)-4ln(2xy)) ?
The answer to (\partial)/(\partial y)(2ln(3x)-4ln(2xy)) is -4/y