What is (\partial)/(\partial x)(4y(1/x-ln(x))) ?

The answer to (\partial)/(\partial x)(4y(1/x-ln(x))) is 4y(-1/(x^2)-1/x)

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(\partial)/(\partial x)(4y(1/x-ln(x)))

\frac{\partial}{\partial x} (4 y (\frac{1}{x} - ln (x)))

4 y (- \frac{1}{x ^{2}} - \frac{1}{x})

Solution steps

\frac{\partial}{\partial x} (4 y (\frac{1}{x} - ln (x)))

Treat

y

as a constant

Take the constant out:

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

= 4 y \frac{\partial}{\partial x} (\frac{1}{x} - ln (x))

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= 4 y (\frac{\partial}{\partial x} (\frac{1}{x}) - \frac{\partial}{\partial x} (ln (x)))

\frac{\partial}{\partial x} (\frac{1}{x}) = - \frac{1}{x ^{2}}

\frac{\partial}{\partial x} (ln (x)) = \frac{1}{x}

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

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What is (\partial)/(\partial x)(4y(1/x-ln(x))) ?
The answer to (\partial)/(\partial x)(4y(1/x-ln(x))) is 4y(-1/(x^2)-1/x)