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

The derivative of y=(2x)/(1-x^2) is (2(1+x^2))/((1-x^2)^2)

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

The first derivative of y=(2x)/(1-x^2) is (2(1+x^2))/((1-x^2)^2)

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derivative of

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

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

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

\frac{d x}{d x} = 1

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

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

Simplify

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

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

\frac{1}{8} x^{\frac{2}{3}} (9 x^{2} - 8 x - 16)

- cos (x)

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

f (x) = x^{2} + 1

x^{2} - 24 x - 12

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