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

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

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

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

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

\int (2 x + 2 y) d y

\frac{d}{d x} (ln (\frac{7 x + 8}{4 x + 8}))

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

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