What is the derivative of ((x+3))/(x^2-4) ?

The derivative of ((x+3))/(x^2-4) is (-x^2-6x-4)/((x^2-4)^2)

What is the first derivative of ((x+3))/(x^2-4) ?

The first derivative of ((x+3))/(x^2-4) is (-x^2-6x-4)/((x^2-4)^2)

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derivative of

\frac{( x + 3 )}{x ^{2} - 4}

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

Solution steps

\frac{d}{d x} (\frac{x + 3}{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 + 3 ) ( x ^{2} - 4 ) - \frac{d}{d x} ( x ^{2} - 4 ) ( x + 3 )}{( x ^{2} - 4 ) ^{2}}

\frac{d}{d x} (x + 3) = 1

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

= \frac{1 \cdot ( x ^{2} - 4 ) - 2 x ( x + 3 )}{( x ^{2} - 4 ) ^{2}}

Simplify

\frac{1 \cdot ( x ^{2} - 4 ) - 2 x ( x + 3 )}{( x ^{2} - 4 ) ^{2}} : \frac{- x ^{2} - 6 x - 4}{( x ^{2} - 4 ) ^{2}}

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

(9 x^{2} + y - 1) d x - (4 y - x) d y = 0

\frac{\partial}{\partial x} (1 + ln (x) + \frac{y}{x})

x \to \infty lim (\frac{ln ( x )}{3 x})

\frac{1}{x - 1}

\frac{\partial}{\partial x} (x^{4} + y^{4} - 4 x + 1 y)

What is the derivative of ((x+3))/(x^2-4) ?
The derivative of ((x+3))/(x^2-4) is (-x^2-6x-4)/((x^2-4)^2)
What is the first derivative of ((x+3))/(x^2-4) ?
The first derivative of ((x+3))/(x^2-4) is (-x^2-6x-4)/((x^2-4)^2)