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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\frac{d y}{d x} + y = e^{x}

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

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

\frac{d}{d x} (sin (k x))

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

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