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

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

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

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

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derivative of

y = \frac{x ^{2} + 1}{x + 1}

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\frac{\partial}{\partial x} (\frac{2 y}{x})

x \to 0 lim (\frac{4}{3 + e ^{\frac{1}{x}}})

\frac{d}{d x} (sin (2 x + 3 y))

\frac{\partial}{\partial y} (ln (x^{2} + y))

a re a x^{4} - 2 x^{2}, 2 x^{2}, [0, 2]

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