What is (\partial)/(\partial x)(4ln(xy)+3ln(yz)+ln(xz)) ?

The answer to (\partial)/(\partial x)(4ln(xy)+3ln(yz)+ln(xz)) is 5/x

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(\partial)/(\partial x)(4ln(xy)+3ln(yz)+ln(xz))

\frac{\partial}{\partial x} (4 ln (x y) + 3 ln (yz) + ln (x z))

\frac{5}{x}

Solution steps

\frac{\partial}{\partial x} (4 ln (x y) + 3 ln (yz) + ln (x z))

Simplify

4 ln (x y) + 3 ln (yz) + ln (x z) : ln (x^{5} y^{7} z^{4})

= \frac{\partial}{\partial x} (ln (x^{5} y^{7} z^{4}))

Treat

y, z

as constants

Apply the chain rule:

\frac{1}{x ^{5} y ^{7} z ^{4}} \frac{\partial}{\partial x} (x^{5} y^{7} z^{4})

= \frac{1}{x ^{5} y ^{7} z ^{4}} \frac{\partial}{\partial x} (x^{5} y^{7} z^{4})

\frac{\partial}{\partial x} (x^{5} y^{7} z^{4}) = 5 y^{7} z^{4} x^{4}

= \frac{1}{x ^{5} y ^{7} z ^{4}} \cdot 5 y^{7} z^{4} x^{4}

Simplify

\frac{1}{x ^{5} y ^{7} z ^{4}} \cdot 5 y^{7} z^{4} x^{4} : \frac{5}{x}

= \frac{5}{x}

x = y^{2} - 4 y, x = y - y^{2}

2 y^{^{''}} - 15 y^{^{'}} - 8 y = 0

\int_{- 1}^{1} x - 1 - x^{2} d x

y = (x^{6} - 10) x

\int sin^{5} (x) cos^{8} (x) d x

What is (\partial)/(\partial x)(4ln(xy)+3ln(yz)+ln(xz)) ?
The answer to (\partial)/(\partial x)(4ln(xy)+3ln(yz)+ln(xz)) is 5/x