What is the derivative of ln(10^{-x}-1) ?

The derivative of ln(10^{-x}-1) is -(ln(10)*10^{-x})/(10^{-x)-1}

What is the first derivative of ln(10^{-x}-1) ?

The first derivative of ln(10^{-x}-1) is -(ln(10)*10^{-x})/(10^{-x)-1}

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\frac{d}{d x} (ln (1 0^{- x} - 1))

- \frac{ln ( 10 ) \cdot 1 0 ^{- x}}{1 0 ^{- x} - 1}

Solution steps

\frac{d}{d x} (ln (1 0^{- x} - 1))

Apply the chain rule:

\frac{1}{1 0 ^{- x} - 1} \frac{d}{d x} (1 0^{- x} - 1)

= \frac{1}{1 0 ^{- x} - 1} \frac{d}{d x} (1 0^{- x} - 1)

\frac{d}{d x} (1 0^{- x} - 1) = - ln (10) \cdot 1 0^{- x}

= \frac{1}{1 0 ^{- x} - 1} (- ln (10) \cdot 1 0^{- x})

Simplify

\frac{1}{1 0 ^{- x} - 1} (- ln (10) \cdot 1 0^{- x}) : - \frac{ln ( 10 ) \cdot 1 0 ^{- x}}{1 0 ^{- x} - 1}

= - \frac{ln ( 10 ) \cdot 1 0 ^{- x}}{1 0 ^{- x} - 1}

\int e^{x^{2}} \cdot x^{3} d x

\frac{d}{d x} ((x^{2} - 3 x + 1)^{4})

5 x, \frac{5}{3} x, 14 - x^{2}

9 e^{x}

y (X) = \frac{1}{2 cos ^{2} ( X )}

What is the derivative of ln(10^{-x}-1) ?
The derivative of ln(10^{-x}-1) is -(ln(10)*10^{-x})/(10^{-x)-1}
What is the first derivative of ln(10^{-x}-1) ?
The first derivative of ln(10^{-x}-1) is -(ln(10)*10^{-x})/(10^{-x)-1}