What is the ((2x+1)^{-1})^' ?

The ((2x+1)^{-1})^' is -2/((2x+1)^2)

What is the first ((2x+1)^{-1})^' ?

The first ((2x+1)^{-1})^' is -2/((2x+1)^2)

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

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

Solution steps

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

Apply the chain rule:

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

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

(2 x + 1)^{'} = 2

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

Simplify

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

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

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

12 x^{2} - 7 x - 4

\frac{d}{d a} ((- 9) - (- 7 a))

y^{^{'}} + x^{- 1} y = x y^{2}

f (x) = e^{x}, at x = 0