What is the derivative of sqrt((1+x/(1-x))) ?

The derivative of sqrt((1+x/(1-x))) is 1/(sqrt(x+1)(-x+1)^{3/2)}

What is the first derivative of sqrt((1+x/(1-x))) ?

The first derivative of sqrt((1+x/(1-x))) is 1/(sqrt(x+1)(-x+1)^{3/2)}

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

(t^{3} - 3 t + 2) e^{- 2 t}

\int_{0}^{5} \frac{1}{1 + x ^{2}} d x

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

x^{2} \frac{d y}{d x} - 2 x y = 6 y^{4}, y (1) = \frac{1}{4}

What is the derivative of sqrt((1+x/(1-x))) ?
The derivative of sqrt((1+x/(1-x))) is 1/(sqrt(x+1)(-x+1)^{3/2)}
What is the first derivative of sqrt((1+x/(1-x))) ?
The first derivative of sqrt((1+x/(1-x))) is 1/(sqrt(x+1)(-x+1)^{3/2)}