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

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

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

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

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

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

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

\frac{d}{d x} ((x^{\frac{1}{2}}) \cdot (x^{2} + x - 4))

\frac{d}{d x} (x^{3} + b x^{2} + c x + d)

\frac{\partial}{\partial y} (sin (x y) + x sin (y))

x \to - 1 - lim ((x + 2 x^{3})^{4})

n = 0 \sum \infty (\frac{5}{5})^{n}

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