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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

y^{^{''}} - 3 y^{^{'}} + 2 y = 2 e^{t}

(y^{2} - 3 x y - 2 x^{2}) d x + (x y - x^{2}) d y = 0

\int cos (a) x d x

x e^{y}

\int (9 - x^{2}) d x

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