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

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

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

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

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derivative of

2 + e^{2 t} + e^{- 2 t}

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{\partial}{\partial x} (36 (\frac{x}{x + y}) - x)

(x^{2} + 1) (\frac{d y}{d x}) + 8 x (y - 1) = 0, y (0) = 6

\frac{\partial}{\partial x} (2 x^{3} y + 1)

x \to 5 lim (x + 2)

x \to 2 lim ((x - 1)^{2} - 1)

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