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

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

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

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

\frac{1}{2 x ^{2}} - \frac{1}{x ^{3}}

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial x} (\frac{1}{2 x ^{2}}) = - \frac{1}{x ^{3}}

= - (- \frac{1}{2 x ^{2}}) - \frac{1}{x ^{3}}

Simplify

= \frac{1}{2 x ^{2}} - \frac{1}{x ^{3}}

\frac{d}{d x} (x^{4} sin (3 x))

\int \frac{6}{x + 1} d x

y = x sin (\frac{3}{x})

\int \frac{x ^{7}}{x ^{4} - 3} d x

(- 7, - 4), (8, 8)

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