What is (\partial)/(\partial x)(ln(xy)+ln(yz)+5ln(xz)) ?

The answer to (\partial)/(\partial x)(ln(xy)+ln(yz)+5ln(xz)) is 6/x

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(\partial)/(\partial x)(ln(xy)+ln(yz)+5ln(xz))

\frac{\partial}{\partial x} (ln (x y) + ln (yz) + 5 ln (x z))

\frac{6}{x}

Solution steps

\frac{\partial}{\partial x} (ln (x y) + ln (yz) + 5 ln (x z))

Simplify

ln (x y) + ln (yz) + 5 ln (x z) : ln (x^{6} y^{2} z^{6})

= \frac{\partial}{\partial x} (ln (x^{6} y^{2} z^{6}))

Treat

y, z

as constants

Apply the chain rule:

\frac{1}{x ^{6} y ^{2} z ^{6}} \frac{\partial}{\partial x} (x^{6} y^{2} z^{6})

= \frac{1}{x ^{6} y ^{2} z ^{6}} \frac{\partial}{\partial x} (x^{6} y^{2} z^{6})

\frac{\partial}{\partial x} (x^{6} y^{2} z^{6}) = 6 y^{2} z^{6} x^{5}

= \frac{1}{x ^{6} y ^{2} z ^{6}} \cdot 6 y^{2} z^{6} x^{5}

Simplify

\frac{1}{x ^{6} y ^{2} z ^{6}} \cdot 6 y^{2} z^{6} x^{5} : \frac{6}{x}

= \frac{6}{x}

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

x^{2} - 4, x + 2

x \to 0 lim (- x ln (x))

\int sin (2 x) - cos (3 x) d x

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

What is (\partial)/(\partial x)(ln(xy)+ln(yz)+5ln(xz)) ?
The answer to (\partial)/(\partial x)(ln(xy)+ln(yz)+5ln(xz)) is 6/x