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

The derivative of ln(x-e^{-x}) is 1/x+e^{-x}

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

The first derivative of ln(x-e^{-x}) is 1/x+e^{-x}

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\frac{d}{d x} (ln (x) - e^{- x})

\frac{1}{x} + e^{- x}

Solution steps

\frac{d}{d x} (ln (x) - e^{- x})

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (ln (x)) - \frac{d}{d x} (e^{- x})

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

\frac{d}{d x} (e^{- x}) = - e^{- x}

= \frac{1}{x} - (- e^{- x})

Simplify

= \frac{1}{x} + e^{- x}

2 x y d y = (x^{2} - y^{2}) d x

\int e^{a x} cos (b x) d x

y = 4 - x^{2}, (- 2, 2)

\frac{\partial}{\partial x} (sin (4 x^{2} y - 4 x y^{3}))

f (x) = \frac{6 x ^{2} + 8 x + 2}{x}

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