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

The derivative of f(x)=(((x+1))/((x-1)))^2 is -(4(x+1))/((x-1)^3)

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

The first derivative of f(x)=(((x+1))/((x-1)))^2 is -(4(x+1))/((x-1)^3)

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derivative of

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

- \frac{4 ( x + 1 )}{( x - 1 ) ^{3}}

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

2 \cdot \frac{x + 1}{x - 1} (- \frac{2}{( x - 1 ) ^{2}}) : - \frac{4 ( x + 1 )}{( x - 1 ) ^{3}}

= - \frac{4 ( x + 1 )}{( x - 1 ) ^{3}}

\int 7 tan^{2} (x) sec (x) d x

x \to 0 lim (\frac{2}{1 + 2 ^{\frac{1}{x}}})

\int \frac{1}{x ^{2} + 16 x} d x

\frac{d}{d x} (\frac{e ^{xx} - e ^{x}}{x ^{2}})

x \to 5 lim (\frac{6}{x})

What is the derivative of f(x)=(((x+1))/((x-1)))^2 ?
The derivative of f(x)=(((x+1))/((x-1)))^2 is -(4(x+1))/((x-1)^3)
What is the first derivative of f(x)=(((x+1))/((x-1)))^2 ?
The first derivative of f(x)=(((x+1))/((x-1)))^2 is -(4(x+1))/((x-1)^3)