What is the derivative of ln(1/4 e^x+3/4 e^{-x}) ?

The derivative of ln(1/4 e^x+3/4 e^{-x}) is (1/4 e^x-3/4 e^{-x})/(1/4 e^x+3/4 e^{-x)}

What is the first derivative of ln(1/4 e^x+3/4 e^{-x}) ?

The first derivative of ln(1/4 e^x+3/4 e^{-x}) is (1/4 e^x-3/4 e^{-x})/(1/4 e^x+3/4 e^{-x)}

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

\frac{\frac{1}{4} e ^{x} - \frac{3}{4} e ^{- x}}{\frac{1}{4} e ^{x} + \frac{3}{4} e ^{- x}}

Solution steps

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

Apply the chain rule:

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

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

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

= \frac{1}{\frac{1}{4} e ^{x} + \frac{3}{4} e ^{- x}} (\frac{1}{4} e^{x} - \frac{3}{4} e^{- x})

Simplify

\frac{1}{\frac{1}{4} e ^{x} + \frac{3}{4} e ^{- x}} (\frac{1}{4} e^{x} - \frac{3}{4} e^{- x}) : \frac{\frac{1}{4} e ^{x} - \frac{3}{4} e ^{- x}}{\frac{1}{4} e ^{x} + \frac{3}{4} e ^{- x}}

= \frac{\frac{1}{4} e ^{x} - \frac{3}{4} e ^{- x}}{\frac{1}{4} e ^{x} + \frac{3}{4} e ^{- x}}

\frac{d}{d x} (ln ∣ x - 1 ∣)

f^{^{'}} (x) = cos (x), f (0) = - 5

\frac{d}{d x} (sin^{2} (- 3 x + 6))

(- 10, 3), (- 8, - 8)

n = 0 \sum \infty (\frac{1}{5})^{n + 1}

What is the derivative of ln(1/4 e^x+3/4 e^{-x}) ?
The derivative of ln(1/4 e^x+3/4 e^{-x}) is (1/4 e^x-3/4 e^{-x})/(1/4 e^x+3/4 e^{-x)}
What is the first derivative of ln(1/4 e^x+3/4 e^{-x}) ?
The first derivative of ln(1/4 e^x+3/4 e^{-x}) is (1/4 e^x-3/4 e^{-x})/(1/4 e^x+3/4 e^{-x)}