What is the derivative of (-2nx^{n-1})/((1+x^n)^2) ?

The derivative of (-2nx^{n-1})/((1+x^n)^2) is -(2nx^{n-2}(-nx^n-x^n+n-1))/((1+x^n)^3)

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

The first derivative of (-2nx^{n-1})/((1+x^n)^2) is -(2nx^{n-2}(-nx^n-x^n+n-1))/((1+x^n)^3)

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derivative of

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

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

Solution steps

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

Simplify

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

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

What is the derivative of (-2nx^{n-1})/((1+x^n)^2) ?
The derivative of (-2nx^{n-1})/((1+x^n)^2) is -(2nx^{n-2}(-nx^n-x^n+n-1))/((1+x^n)^3)
What is the first derivative of (-2nx^{n-1})/((1+x^n)^2) ?
The first derivative of (-2nx^{n-1})/((1+x^n)^2) is -(2nx^{n-2}(-nx^n-x^n+n-1))/((1+x^n)^3)