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

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

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

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

Simplify

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

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

\int (- 2 cos (x)) d x

\int sin (\frac{t}{2}) t d t

1 - x^{2}, - 1, 1

\frac{d}{d a} (tan (a) \cdot cot (a))

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 9 y = sec (3 x)

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