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

The derivative of (x/(2(x^2-1)^2-1))^2 is (2x(-6x^4+4x^2+1))/((2x^4-4x^2+1)^3)

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

The first derivative of (x/(2(x^2-1)^2-1))^2 is (2x(-6x^4+4x^2+1))/((2x^4-4x^2+1)^3)

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derivative of

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

\frac{2 x ( - 6 x ^{4} + 4 x ^{2} + 1 )}{( 2 x ^{4} - 4 x ^{2} + 1 ) ^{3}}

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (\frac{x}{2 ( x ^{2} - 1 ) ^{2} - 1}) = \frac{- 6 x ^{4} + 4 x ^{2} + 1}{( 2 x ^{4} - 4 x ^{2} + 1 ) ^{2}}

= 2 \cdot \frac{x}{2 ( x ^{2} - 1 ) ^{2} - 1} \cdot \frac{- 6 x ^{4} + 4 x ^{2} + 1}{( 2 x ^{4} - 4 x ^{2} + 1 ) ^{2}}

Simplify

2 \cdot \frac{x}{2 ( x ^{2} - 1 ) ^{2} - 1} \cdot \frac{- 6 x ^{4} + 4 x ^{2} + 1}{( 2 x ^{4} - 4 x ^{2} + 1 ) ^{2}} : \frac{2 x ( - 6 x ^{4} + 4 x ^{2} + 1 )}{( 2 x ^{4} - 4 x ^{2} + 1 ) ^{3}}

= \frac{2 x ( - 6 x ^{4} + 4 x ^{2} + 1 )}{( 2 x ^{4} - 4 x ^{2} + 1 ) ^{3}}

\int_{0}^{30} 4 x (30 - x) d x

\frac{\partial}{\partial x} (x^{3} e^{4 x y})

y^{^{'}} = \frac{y}{t} - \frac{1}{\frac{y}{t}}

f (x) = x + \frac{2}{x}, at x = 2

\int \frac{50}{x ^{0.25}} d x

What is the derivative of (x/(2(x^2-1)^2-1))^2 ?
The derivative of (x/(2(x^2-1)^2-1))^2 is (2x(-6x^4+4x^2+1))/((2x^4-4x^2+1)^3)
What is the first derivative of (x/(2(x^2-1)^2-1))^2 ?
The first derivative of (x/(2(x^2-1)^2-1))^2 is (2x(-6x^4+4x^2+1))/((2x^4-4x^2+1)^3)