What is the derivative of (6x/(x^2-4)) ?

The derivative of (6x/(x^2-4)) is (6(-x^2-4))/((x^2-4)^2)

What is the first derivative of (6x/(x^2-4)) ?

The first derivative of (6x/(x^2-4)) is (6(-x^2-4))/((x^2-4)^2)

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\frac{d}{d x} (\frac{6 x}{x ^{2} - 4})

\frac{6 ( - x ^{2} - 4 )}{( x ^{2} - 4 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{6 x}{x ^{2} - 4})

Take the constant out:

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

= 6 \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}}

= 6 \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

= 6 \cdot \frac{1 \cdot ( x ^{2} - 4 ) - 2 xx}{( x ^{2} - 4 ) ^{2}}

Simplify

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

= \frac{6 ( - x ^{2} - 4 )}{( x ^{2} - 4 ) ^{2}}

\int_{0}^{N} e^{- s t} e^{- 4 t} d t

x \to + 0 + lim (\frac{5 x}{sin ( x )})

\frac{1}{x} \cdot 3 e^{x}

\int \frac{1}{9 x ^{2} - 4} d x

\int \frac{1}{2 x + x ^{3}} d x

What is the derivative of (6x/(x^2-4)) ?
The derivative of (6x/(x^2-4)) is (6(-x^2-4))/((x^2-4)^2)
What is the first derivative of (6x/(x^2-4)) ?
The first derivative of (6x/(x^2-4)) is (6(-x^2-4))/((x^2-4)^2)