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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

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

\frac{d}{d x} ((x + 1)^{2} (x - 1)^{2}) = 4 x^{3} - 4 x

= - 2 \cdot \frac{1 \cdot ( x + 1 ) ^{2} ( x - 1 ) ^{2} - ( 4 x ^{3} - 4 x ) x}{( ( x + 1 ) ^{2} ( x - 1 ) ^{2} ) ^{2}}

Simplify

- 2 \cdot \frac{1 \cdot ( x + 1 ) ^{2} ( x - 1 ) ^{2} - ( 4 x ^{3} - 4 x ) x}{( ( x + 1 ) ^{2} ( x - 1 ) ^{2} ) ^{2}} : \frac{2 ( 3 x ^{2} + 1 )}{( x + 1 ) ^{3} ( x - 1 ) ^{3}}

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

\int [5 x^{4} + 10 cos (x)] d x

\int \frac{5 + x ^{3} + 2 x}{x} d x

\frac{\partial}{\partial x} (3 (x - 6)^{3})

3 - 11 x^{5}

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

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