What is the derivative of y=sqrt(e^{2x)+e^{-2x}} ?

The derivative of y=sqrt(e^{2x)+e^{-2x}} is (e^{-2x}(e^{4x}-1))/(sqrt(e^{2x)+e^{-2x)}}

What is the first derivative of y=sqrt(e^{2x)+e^{-2x}} ?

The first derivative of y=sqrt(e^{2x)+e^{-2x}} is (e^{-2x}(e^{4x}-1))/(sqrt(e^{2x)+e^{-2x)}}

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derivative of

y = e^{2 x} + e^{- 2 x}

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

\frac{1}{2 e ^{2 x} + e ^{- 2 x}} (e^{2 x} \cdot 2 - 2 e^{- 2 x}) : \frac{e ^{- 2 x} ( e ^{4 x} - 1 )}{e ^{2 x} + e ^{- 2 x}}

= \frac{e ^{- 2 x} ( e ^{4 x} - 1 )}{e ^{2 x} + e ^{- 2 x}}

\frac{d y}{d x} = x + y, y (0) = 0

f (x) = 3 e^{- 2 x} ln (x)

\frac{d}{d x} (arcsin (5 x^{3}))

\frac{d}{d x} (ln ∣ csc (3 x) ∣)

\int 4 e^{- 0.4 x} d x

What is the derivative of y=sqrt(e^{2x)+e^{-2x}} ?
The derivative of y=sqrt(e^{2x)+e^{-2x}} is (e^{-2x}(e^{4x}-1))/(sqrt(e^{2x)+e^{-2x)}}
What is the first derivative of y=sqrt(e^{2x)+e^{-2x}} ?
The first derivative of y=sqrt(e^{2x)+e^{-2x}} is (e^{-2x}(e^{4x}-1))/(sqrt(e^{2x)+e^{-2x)}}