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

The derivative of y=((3x-6))/((x^2+4)) is (-3x^2+12x+12)/((x^2+4)^2)

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

The first derivative of y=((3x-6))/((x^2+4)) is (-3x^2+12x+12)/((x^2+4)^2)

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derivative of

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

\frac{- 3 x ^{2} + 12 x + 12}{( x ^{2} + 4 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Expand

3 (x^{2} + 4) - 2 x (3 x - 6) : - 3 x^{2} + 12 x + 12

= \frac{- 3 x ^{2} + 12 x + 12}{( x ^{2} + 4 ) ^{2}}

x^{2} + y^{3} = 12

\frac{d}{d x} (9 e^{x} + 5)

\frac{16}{( s + 4 ) ^{2}}

\frac{d y}{d x} - 5 y = 0

\int \frac{1}{sin ^{2} ( u )} d u

What is the derivative of y=((3x-6))/((x^2+4)) ?
The derivative of y=((3x-6))/((x^2+4)) is (-3x^2+12x+12)/((x^2+4)^2)
What is the first derivative of y=((3x-6))/((x^2+4)) ?
The first derivative of y=((3x-6))/((x^2+4)) is (-3x^2+12x+12)/((x^2+4)^2)