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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} ((x - 1) (x + 1)^{2}) = 3 x^{2} + 2 x - 1

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

Simplify

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

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

\int_{1}^{x^{3}} t^{2} d t

x \to 0 lim (\frac{e ^{- 5 x} - 1}{x})

\frac{d}{d x} (cos (5 x^{2} - x))

f (x) = 2 x^{2} + 3 x, at x = - 2

f (x) = e^{3 e^{x}}

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