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

The derivative of ((-2x/(1+x^2))^2) is (8x(1-x^2))/((1+x^2)^3)

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

The first derivative of ((-2x/(1+x^2))^2) is (8x(1-x^2))/((1+x^2)^3)

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

\frac{8 x ( 1 - x ^{2} )}{( 1 + x ^{2} ) ^{3}}

Solution steps

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

Apply the chain rule:

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

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

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

Simplify

2 \cdot \frac{- 2 x}{1 + x ^{2}} (- \frac{2 ( 1 - x ^{2} )}{( 1 + x ^{2} ) ^{2}}) : \frac{8 x ( 1 - x ^{2} )}{( 1 + x ^{2} ) ^{3}}

= \frac{8 x ( 1 - x ^{2} )}{( 1 + x ^{2} ) ^{3}}

2 x y y^{^{'}} = 1 + y^{2}

x \to 0 lim (- 5 (x - 3) ln (x - 3))

y^{^{'}} = 2 (y^{2} + 1)

\frac{d}{d x} (x^{2} - x y) = 7

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

What is the derivative of ((-2x/(1+x^2))^2) ?
The derivative of ((-2x/(1+x^2))^2) is (8x(1-x^2))/((1+x^2)^3)
What is the first derivative of ((-2x/(1+x^2))^2) ?
The first derivative of ((-2x/(1+x^2))^2) is (8x(1-x^2))/((1+x^2)^3)