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

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\frac{sec ( x )}{1 + sec ( x )}

\int x ln^{3} (x) d x

\int_{0}^{2} \frac{1}{4 + x ^{2}} d x

\frac{d}{d x} (a x)

n = 1 \sum \infty n sin (nπ)

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