What is (\partial)/(\partial x)(4ln((2xy)/(7z))) ?

The answer to (\partial)/(\partial x)(4ln((2xy)/(7z))) is 4/x

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(\partial)/(\partial x)(4ln((2xy)/(7z)))

\frac{\partial}{\partial x} (4 ln (\frac{2 x y}{7 z}))

\frac{4}{x}

Solution steps

\frac{\partial}{\partial x} (4 ln (\frac{2 x y}{7 z}))

Treat

y, z

as constants

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 4 \frac{\partial}{\partial x} (ln (\frac{2 x y}{7 z}))

Apply the chain rule:

\frac{1}{\frac{2 x y}{7 z}} \frac{\partial}{\partial x} (\frac{2 x y}{7 z})

= \frac{1}{\frac{2 x y}{7 z}} \frac{\partial}{\partial x} (\frac{2 x y}{7 z})

\frac{\partial}{\partial x} (\frac{2 x y}{7 z}) = \frac{2 y}{7 z}

= 4 \cdot \frac{1}{\frac{2 x y}{7 z}} \cdot \frac{2 y}{7 z}

Simplify

4 \cdot \frac{1}{\frac{2 x y}{7 z}} \cdot \frac{2 y}{7 z} : \frac{4}{x}

= \frac{4}{x}

h (w) = \frac{4 w ^{5} - w}{w}

\int x e^{1 - x} d x

f (x) = x sin (\frac{1}{x})

y = x^{2} - 3 x - 2, (2, - 4)

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

What is (\partial)/(\partial x)(4ln((2xy)/(7z))) ?
The answer to (\partial)/(\partial x)(4ln((2xy)/(7z))) is 4/x