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

The derivative of f(x)=(x-1)/(x^2+4) is (-x^2+2x+4)/((x^2+4)^2)

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

The first derivative of f(x)=(x-1)/(x^2+4) is (-x^2+2x+4)/((x^2+4)^2)

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derivative of

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

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

Solution steps

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

\frac{d}{d x} (x - 1) = 1

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

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

Simplify

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

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

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