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

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

z^{3} - 5 z + 4

y = \frac{x + 2}{x - 2}

\int cos (10 x) d x

d y = (9 x + 4 y + 1)^{2} d x

\int tan^{2} (x) sec^{3} (x) d x

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