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

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

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

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

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derivative of

f (x) = \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}}

\int (7^{x}) d x

\int x^{3} e^{\frac{x ^{2}}{2}} d x

\int \frac{4}{x ^{2} + 5 x - 4} d x

n = 1 \sum \infty \frac{9}{n ( n + 3 )}

\frac{- 1}{x}

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