What is the derivative of (x+1)^2(x^2+1)^{-3} ?

The derivative of (x+1)^2(x^2+1)^{-3} is (-4x^3-10x^2-4x+2)/((x^2+1)^4)

What is the first derivative of (x+1)^2(x^2+1)^{-3} ?

The first derivative of (x+1)^2(x^2+1)^{-3} is (-4x^3-10x^2-4x+2)/((x^2+1)^4)

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derivative of

(x + 1)^{2} (x^{2} + 1)^{- 3}

\frac{- 4 x ^{3} - 10 x ^{2} - 4 x + 2}{( x ^{2} + 1 ) ^{4}}

Solution steps

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = (x + 1)^{2}, g = (x^{2} + 1)^{- 3}

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

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

\frac{d}{d x} ((x^{2} + 1)^{- 3}) = - \frac{6 x}{( x ^{2} + 1 ) ^{4}}

= 2 (x + 1) (x^{2} + 1)^{- 3} + (- \frac{6 x}{( x ^{2} + 1 ) ^{4}}) (x + 1)^{2}

Simplify

2 (x + 1) (x^{2} + 1)^{- 3} + (- \frac{6 x}{( x ^{2} + 1 ) ^{4}}) (x + 1)^{2} : \frac{- 4 x ^{3} - 10 x ^{2} - 4 x + 2}{( x ^{2} + 1 ) ^{4}}

= \frac{- 4 x ^{3} - 10 x ^{2} - 4 x + 2}{( x ^{2} + 1 ) ^{4}}

\frac{\partial}{\partial x} (a cos (x))

x \to 0 + lim (\frac{x - 2}{∣ x ∣})

T

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

x \to 0 - lim (2 x - 8 x^{2})

What is the derivative of (x+1)^2(x^2+1)^{-3} ?
The derivative of (x+1)^2(x^2+1)^{-3} is (-4x^3-10x^2-4x+2)/((x^2+1)^4)
What is the first derivative of (x+1)^2(x^2+1)^{-3} ?
The first derivative of (x+1)^2(x^2+1)^{-3} is (-4x^3-10x^2-4x+2)/((x^2+1)^4)