What is the derivative of (3-4x/(x^2+1)) ?

The derivative of (3-4x/(x^2+1)) is (4x^2-6x-4)/((x^2+1)^2)

What is the first derivative of (3-4x/(x^2+1)) ?

The first derivative of (3-4x/(x^2+1)) is (4x^2-6x-4)/((x^2+1)^2)

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (3 - 4 x) = - 4

\frac{d}{d x} (x^{2} + 1) = 2 x

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

Simplify

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

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

What is the derivative of (3-4x/(x^2+1)) ?
The derivative of (3-4x/(x^2+1)) is (4x^2-6x-4)/((x^2+1)^2)
What is the first derivative of (3-4x/(x^2+1)) ?
The first derivative of (3-4x/(x^2+1)) is (4x^2-6x-4)/((x^2+1)^2)