What is (\partial)/(\partial x)((1-ln(x))/(x^2)) ?

The answer to (\partial)/(\partial x)((1-ln(x))/(x^2)) is -(-2ln(x)+3)/(x^3)

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(\partial)/(\partial x)((1-ln(x))/(x^2))

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

- \frac{- 2 ln ( x ) + 3}{x ^{3}}

Solution steps

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

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

Simplify

\frac{( - \frac{1}{x} ) x ^{2} - 2 x ( 1 - ln ( x ) )}{( x ^{2} ) ^{2}} : - \frac{- 2 ln ( x ) + 3}{x ^{3}}

= - \frac{- 2 ln ( x ) + 3}{x ^{3}}

\frac{\partial}{\partial x} (ln (9 x))

\frac{d}{d x} (\frac{4}{7 x ^{4}})

y^{^{'}} + sin (t) y = sin (t), y (0) = 7

y = x^{2} + 9 x, at x = - 8

\frac{d}{d x} (- cos (e^{x}))

What is (\partial)/(\partial x)((1-ln(x))/(x^2)) ?
The answer to (\partial)/(\partial x)((1-ln(x))/(x^2)) is -(-2ln(x)+3)/(x^3)