What is the derivative of g(x)= 1/(x+1) ?

The derivative of g(x)= 1/(x+1) is -1/((x+1)^2)

What is the first derivative of g(x)= 1/(x+1) ?

The first derivative of g(x)= 1/(x+1) is -1/((x+1)^2)

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derivative of

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

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

Solution steps

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= \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}}

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