What is the derivative of ln(3x-e^{-2x}) ?

The derivative of ln(3x-e^{-2x}) is 1/x+2e^{-2x}

What is the first derivative of ln(3x-e^{-2x}) ?

The first derivative of ln(3x-e^{-2x}) is 1/x+2e^{-2x}

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\frac{d}{d x} (ln (3 x) - e^{- 2 x})

\frac{1}{x} + 2 e^{- 2 x}

Solution steps

\frac{d}{d x} (ln (3 x) - e^{- 2 x})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (ln (3 x)) - \frac{d}{d x} (e^{- 2 x})

\frac{d}{d x} (ln (3 x)) = \frac{1}{x}

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

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

Simplify

= \frac{1}{x} + 2 e^{- 2 x}

\int (cot (b x) + tan (b x))^{2} d x

f (x) = x^{2}, at x = 2 x - 1

\frac{d}{d y} (1 - y^{2})

n = 0 \sum \infty (- \frac{4}{9})^{n}

y = x^{\frac{1}{2}}, x = 0, x = 1

What is the derivative of ln(3x-e^{-2x}) ?
The derivative of ln(3x-e^{-2x}) is 1/x+2e^{-2x}
What is the first derivative of ln(3x-e^{-2x}) ?
The first derivative of ln(3x-e^{-2x}) is 1/x+2e^{-2x}