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

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

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

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

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

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

Solution steps

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

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

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

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

Simplify

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

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

\int_{1}^{e^{3}} \frac{ln ^{5} ( x ^{2} )}{x} d x

35 cos (x)

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

x \to - 1 lim (\frac{x ^{2} - 2}{1 - x})

y^{^{'}} = 4 y^{2}

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