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

The derivative of y= 2/(1-x^2) is (4x)/((1-x^2)^2)

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

The first derivative of y= 2/(1-x^2) is (4x)/((1-x^2)^2)

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derivative of

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

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

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

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

x \to 3 lim (x 25 - x^{2})

\frac{\partial}{\partial x} (e^{y})

\int 2 sin (4 x) d x

f (x) = x - 4

What is the derivative of y= 2/(1-x^2) ?
The derivative of y= 2/(1-x^2) is (4x)/((1-x^2)^2)
What is the first derivative of y= 2/(1-x^2) ?
The first derivative of y= 2/(1-x^2) is (4x)/((1-x^2)^2)