What is the derivative of (6x/((x^2-1)^2)) ?

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

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

\int_{4}^{9} x d x

f (x) = sin (x) + x cos (x)

f (x) = x + 3

t y^{^{'}} + 4 y = t^{2} - t + 1

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