What is the derivative of (2x/((x^2-3)^2)) ?

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

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

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

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

Simplify

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

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

d er i v a t i v e y = e^{- x} cos (2 x)

in v erse l a pl a ce \frac{1}{s ^{4} + s ^{2}}

\int (x^{2} + 2 x) d x

\int - 3 tan (3 y) d y

\frac{d}{d x} (\frac{x - y}{sin ( x ) - sin ( y )})

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