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

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

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

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

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derivative

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} (x^{2} + 2) = 2 x

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

Expand

3 (x^{2} + 2) - 2 x (3 x - 2) : - 3 x^{2} + 4 x + 6

= \frac{- 3 x ^{2} + 4 x + 6}{( x ^{2} + 2 ) ^{2}}

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