What is the derivative of 8/(1+x) ?

The derivative of 8/(1+x) is -8/((1+x)^2)

What is the first derivative of 8/(1+x) ?

The first derivative of 8/(1+x) is -8/((1+x)^2)

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\frac{d}{d x} (\frac{8}{1 + x})

- \frac{8}{( 1 + x ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{8}{1 + x})

Take the constant out:

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

= 8 \frac{d}{d x} (\frac{1}{1 + x})

Apply exponent rule:

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

= 8 \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

= 8 (- \frac{1}{( 1 + x ) ^{2}} \cdot 1)

Simplify

8 (- \frac{1}{( 1 + x ) ^{2}} \cdot 1) : - \frac{8}{( 1 + x ) ^{2}}

= - \frac{8}{( 1 + x ) ^{2}}

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What is the derivative of 8/(1+x) ?
The derivative of 8/(1+x) is -8/((1+x)^2)
What is the first derivative of 8/(1+x) ?
The first derivative of 8/(1+x) is -8/((1+x)^2)