How do you differentiate $y = {sec}^{- 1} (e^{2 x})$ $y=sec^-1(e^(2x))$ ?

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How do you differentiate $y = {sec}^{- 1} (e^{2 x})$ $y=sec^-1(e^(2x))$ ?

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Answer 1 · 2018-02-18T03:33:17

$\frac{2}{\sqrt{e^{4 x} - 1}}$ $2/sqrt{e^{4x}-1}$

Explanation:

$sec y = e^{2 x}$ $sec y = e^{2x}$

differentiating both sides with respect to $x$ $x$ :

$sec y tan y \frac{d y}{d x} = 2 e^{2 x} \Rightarrow$ $sec y tan y {dy}/{dx}=2e^{2x} implies$

$\frac{d y}{d x} = \frac{2 e^{2 x}}{sec y tan y} = \frac{2 e^{2 x}}{e^{2 x} \sqrt{{(e^{2 x})}^{2} - 1}} = \frac{2}{\sqrt{e^{4 x} - 1}}$ $dy/dx = {2 e^{2x}}/{sec y tan y}={2e^{2x}}/{e^{2x}sqrt{(e^{2x})^2-1}}=2/sqrt{e^{4x}-1}$

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How do you differentiate $y = {sec}^{- 1} (e^{2 x})$ $y=sec^-1(e^(2x))$ ?

1 Answer

Explanation:

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Science

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Humanities

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How do you differentiate y=sec^-1(e^(2x))y=sec−1(e2x)y=sec^-1(e^(2x))?

1 Answer

Explanation:

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How do you differentiate $y = {sec}^{- 1} (e^{2 x})$ $y=sec^-1(e^(2x))$ ?