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

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

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

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

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

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

Solution steps

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

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

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

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

Simplify

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

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

\frac{d}{d x} ((3 x + 5) x^{2})

y^{^{''}} - 6 y^{^{'}} + 8 y = 0

\int sin^{7} (x) cos^{2} (x) d x

y = lo g_{5} (x)

\frac{d}{d x} (lo g_{10} (2 x^{2} + 5 x + 2))

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