What is the derivative of arctan((2x/(1+x^2))) ?

The derivative of arctan((2x/(1+x^2))) is (2(1-x^2))/(x^4+6x^2+1)

What is the first derivative of arctan((2x/(1+x^2))) ?

The first derivative of arctan((2x/(1+x^2))) is (2(1-x^2))/(x^4+6x^2+1)

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

\frac{2 ( 1 - x ^{2} )}{x ^{4} + 6 x ^{2} + 1}

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

\frac{1}{( \frac{2 x}{1 + x ^{2}} ) ^{2} + 1} \cdot \frac{2 ( 1 - x ^{2} )}{( 1 + x ^{2} ) ^{2}} : \frac{2 ( 1 - x ^{2} )}{x ^{4} + 6 x ^{2} + 1}

= \frac{2 ( 1 - x ^{2} )}{x ^{4} + 6 x ^{2} + 1}

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

f (x) = \frac{1}{ln ( x )}

\frac{d}{d x} (\frac{3 - x e ^{x}}{x + e ^{x}})

\frac{d}{d x} (cos (2 x) sin (3 x))

What is the derivative of arctan((2x/(1+x^2))) ?
The derivative of arctan((2x/(1+x^2))) is (2(1-x^2))/(x^4+6x^2+1)
What is the first derivative of arctan((2x/(1+x^2))) ?
The first derivative of arctan((2x/(1+x^2))) is (2(1-x^2))/(x^4+6x^2+1)