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

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

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

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

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\frac{d}{d x} (- \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}}

f (x) = - x^{2} sin (x) + 4 x cos (x) + 2 sin (x)

\frac{d}{d x} (\frac{3}{x ^{4}})

\frac{\partial}{\partial x} (7 - x^{5} - y^{3})

\frac{\partial}{\partial u} (x^{5} v^{2} + 2 y^{3} u - 3)

\frac{\partial}{\partial x} (61 - x^{2} - 3 y^{2})

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