What is the derivative of (x^4-16x^2-16)/((x^2-4)^2) ?

The derivative of (x^4-16x^2-16)/((x^2-4)^2) is (16x(x^2+12))/((x^2-4)^3)

What is the first derivative of (x^4-16x^2-16)/((x^2-4)^2) ?

The first derivative of (x^4-16x^2-16)/((x^2-4)^2) is (16x(x^2+12))/((x^2-4)^3)

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derivative of

\frac{x ^{4} - 16 x ^{2} - 16}{( x ^{2} - 4 ) ^{2}}

\frac{16 x ( x ^{2} + 12 )}{( x ^{2} - 4 ) ^{3}}

Solution steps

\frac{d}{d x} (\frac{x ^{4} - 16 x ^{2} - 16}{( x ^{2} - 4 ) ^{2}})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{4} - 16 x ^{2} - 16 ) ( x ^{2} - 4 ) ^{2} - \frac{d}{d x} ( ( x ^{2} - 4 ) ^{2} ) ( x ^{4} - 16 x ^{2} - 16 )}{( ( x ^{2} - 4 ) ^{2} ) ^{2}}

\frac{d}{d x} (x^{4} - 16 x^{2} - 16) = 4 x^{3} - 32 x

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

= \frac{( 4 x ^{3} - 32 x ) ( x ^{2} - 4 ) ^{2} - 4 x ( x ^{2} - 4 ) ( x ^{4} - 16 x ^{2} - 16 )}{( ( x ^{2} - 4 ) ^{2} ) ^{2}}

Simplify

\frac{( 4 x ^{3} - 32 x ) ( x ^{2} - 4 ) ^{2} - 4 x ( x ^{2} - 4 ) ( x ^{4} - 16 x ^{2} - 16 )}{( ( x ^{2} - 4 ) ^{2} ) ^{2}} : \frac{16 x ( x ^{2} + 12 )}{( x ^{2} - 4 ) ^{3}}

= \frac{16 x ( x ^{2} + 12 )}{( x ^{2} - 4 ) ^{3}}

x y d y - (x^{2} + y^{2}) d x = 0

n = 1 \sum \infty \frac{1}{( 0.8 ) ^{n}}

\int_{0}^{2} x^{2} 4 - x^{2} d x

f (x) = \frac{- 8 x}{x ^{2} + 1}

3 t + 2

What is the derivative of (x^4-16x^2-16)/((x^2-4)^2) ?
The derivative of (x^4-16x^2-16)/((x^2-4)^2) is (16x(x^2+12))/((x^2-4)^3)
What is the first derivative of (x^4-16x^2-16)/((x^2-4)^2) ?
The first derivative of (x^4-16x^2-16)/((x^2-4)^2) is (16x(x^2+12))/((x^2-4)^3)