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

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

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

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

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derivative of

f (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}}

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

f (x) = 5 x^{2} - 8 x + 3, at x = 1

x \to 0 lim (tan (\frac{x}{2}))

x arcsin (x)

n = 1 \sum \infty n!^{2} (5^{n})

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