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

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

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

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

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derivative of

y = \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}}

t an g e n t f (x) = 1260 - 16 x^{2}, at x = 3

d er i v a t i v e f (x) = x^{\frac{19}{9}} + x^{\frac{10}{9}}

\int x z d y

\frac{d}{d x} (15 x^{4} - 15 x^{2})

d er i v a t i v e (u^{2} + u)^{3}

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