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

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\frac{16}{( 1 - x ) ^{3}}

\int \frac{( 9 - x ^{2} )}{x ^{2}} d x

\int \frac{x}{( x ^{2} + 8 ) ^{2}} d x

f (x) = x^{- 12}

\frac{d}{d x} (4 x^{3} + 4 x^{2} - 8 x)

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