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

The derivative of-8/(1-8x) is -(64)/((1-8x)^2)

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

The first derivative of-8/(1-8x) is -(64)/((1-8x)^2)

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

- \frac{64}{( 1 - 8 x ) ^{2}}

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

- 8 (- \frac{1}{( 1 - 8 x ) ^{2}} (- 8)) : - \frac{64}{( 1 - 8 x ) ^{2}}

= - \frac{64}{( 1 - 8 x ) ^{2}}

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

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\int \frac{ln ( 2 x )}{2 x ^{2}} d x

What is the derivative of-8/(1-8x) ?
The derivative of-8/(1-8x) is -(64)/((1-8x)^2)
What is the first derivative of-8/(1-8x) ?
The first derivative of-8/(1-8x) is -(64)/((1-8x)^2)