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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Expand

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

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

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