What is (\partial)/(\partial y)(3ln(x)+ln(y)) ?

The answer to (\partial)/(\partial y)(3ln(x)+ln(y)) is 1/y

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(\partial)/(\partial y)(3ln(x)+ln(y))

\frac{\partial}{\partial y} (3 ln (x) + ln (y))

\frac{1}{y}

Solution steps

\frac{\partial}{\partial y} (3 ln (x) + ln (y))

Simplify

3 ln (x) + ln (y) : ln (x^{3} y)

= \frac{\partial}{\partial y} (ln (x^{3} y))

Treat

x

as a constant

Apply the chain rule:

\frac{1}{x ^{3} y} \frac{\partial}{\partial y} (x^{3} y)

= \frac{1}{x ^{3} y} \frac{\partial}{\partial y} (x^{3} y)

\frac{\partial}{\partial y} (x^{3} y) = x^{3}

= \frac{1}{x ^{3} y} x^{3}

Simplify

\frac{1}{x ^{3} y} x^{3} : \frac{1}{y}

= \frac{1}{y}

y^{^{''}} + 16 y^{^{'}} = 0, y (0) = 2, y (0) = - 2

x \to 1 lim (3 x + 7)

y = 4 x + 84, (4, 10)

\frac{d}{d x} (3 x^{2} (2 x + 5)^{3})

\int \frac{cos ( \frac{π}{x ^{9}} )}{x ^{10}} d x

What is (\partial)/(\partial y)(3ln(x)+ln(y)) ?
The answer to (\partial)/(\partial y)(3ln(x)+ln(y)) is 1/y