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

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

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

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

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Popular >

\frac{d}{d x} (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}}

(x - 3)^{^{'}}

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

\int \frac{3 + 4 x}{1 + x ^{2}} d x

\int_{0}^{\frac{π}{4}} x^{2} cos (2 x) d x

f (x) = cos (8 x)

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