What is the derivative of (6x/(x^2-3)) ?

The derivative of (6x/(x^2-3)) is (6(-x^2-3))/((x^2-3)^2)

What is the first derivative of (6x/(x^2-3)) ?

The first derivative of (6x/(x^2-3)) is (6(-x^2-3))/((x^2-3)^2)

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\frac{d}{d x} (\frac{6 x}{x ^{2} - 3})

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

Solution steps

\frac{d}{d x} (\frac{6 x}{x ^{2} - 3})

Take the constant out:

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

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

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

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

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

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

Simplify

6 \cdot \frac{1 \cdot ( x ^{2} - 3 ) - 2 xx}{( x ^{2} - 3 ) ^{2}} : \frac{6 ( - x ^{2} - 3 )}{( x ^{2} - 3 ) ^{2}}

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

x \to - 5 lim (\frac{x ^{3} + 125}{x ^{2} - 25})

\int \frac{x - 2}{x + 7} d x

\int_{0}^{9} x^{2} - 6 x + 5 d x

x \to 1 - lim (\frac{∣ x - 1 ∣}{1 - x})

y = 8 e^{8 e^{x} + x}

What is the derivative of (6x/(x^2-3)) ?
The derivative of (6x/(x^2-3)) is (6(-x^2-3))/((x^2-3)^2)
What is the first derivative of (6x/(x^2-3)) ?
The first derivative of (6x/(x^2-3)) is (6(-x^2-3))/((x^2-3)^2)