Question

Differentiate
e
${e}^{2x} + {e}^{ - 2x} \div {e}^{2x} - {e}^{ - 2x}$
​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

$\tt\implies \:Let \: y \: = \: \dfrac{{e}^{2x} + {e}^{ - 2x}}{{e}^{2x} - {e}^{ - 2x}}$

$\begin{gathered}\Large{\bold{\pink{\underline{Formula \: Used \::}}}} \end{gathered}$

$\tt\implies \:\dfrac{d}{dx} {e}^{x} = {e}^{x}$

$\tt\implies \:\dfrac{d}{dx} ( \dfrac{u}{v} ) = \dfrac{v\dfrac{d}{dx} u - u\dfrac{d}{dx} v}{ {v}^{2} }$

$\tt\implies \: {(x + y)}^{2} - {(x - y)}^{2} = 4xy$

$\large\underline\purple{\bold{Solution :- }}$

$\tt\implies \ \: y \: = \: \dfrac{{e}^{2x} + {e}^{ - 2x}}{{e}^{2x} - {e}^{ - 2x}}$

☆ On differentiating both sides w. r. t. x, we get

$\tt\implies \: \dfrac{d}{dx} \: y \: = \dfrac{d}{dx} \: \dfrac{{e}^{2x} + {e}^{ - 2x}}{{e}^{2x} - {e}^{ - 2x}}$

$\tt\dfrac{dy}{dx} = \dfrac{({e}^{2x} - {e}^{ - 2x})\dfrac{d}{dx} ({e}^{2x} + {e}^{ - 2x}) - ({e}^{2x} + {e}^{ - 2x})\dfrac{d}{dx} ({e}^{2x} - {e}^{ - 2x})}{ {({e}^{2x} - {e}^{ - 2x})}^{2} }$

$\tt \: \dfrac{dy}{dx} = \dfrac{({e}^{2x} - {e}^{ - 2x})(2{e}^{2x} - 2 {e}^{ - 2x}) - ({e}^{2x} + {e}^{ - 2x})(2{e}^{2x} + 2{e}^{ - 2x})}{ {({e}^{2x} - {e}^{ - 2x})}^{2} }$

$\tt \:\dfrac{dy}{dx} = \dfrac{2({e}^{2x} - {e}^{ - 2x})({e}^{2x} - {e}^{ - 2x}) -2 ({e}^{2x} + {e}^{ - 2x})({e}^{2x} + {e}^{ - 2x})}{ {({e}^{2x} - {e}^{ - 2x})}^{2} }$

$\tt \:\dfrac{dy}{dx} = \dfrac{2 \bigg(({e}^{2x} - {e}^{ - 2x})({e}^{2x} - {e}^{ - 2x}) - ({e}^{2x} + {e}^{ - 2x})({e}^{2x} + {e}^{ - 2x}) \bigg)}{ {({e}^{2x} - {e}^{ - 2x})}^{2} }$

$\tt\implies \:\dfrac{dy}{dx} = \dfrac{2 \bigg( {({e}^{2x} - {e}^{ - 2x})}^{2} - {({e}^{2x} + {e}^{ - 2x})}^{2} \bigg) }{ {({e}^{2x} - {e}^{ - 2x})}^{2} }$

$\tt\implies \:\dfrac{dy}{dx} = \dfrac{ - 2 \bigg( {({e}^{2x} + {e}^{ - 2x})}^{2} - {({e}^{2x} - {e}^{ - 2x})}^{2} \bigg) }{ {({e}^{2x} - {e}^{ - 2x})}^{2} }$

$\tt\implies \:\dfrac{dy}{dx} = \dfrac{ - 2(4 \times {e}^{2x} \times {e}^{ - 2x})}{ {({e}^{2x} - {e}^{ - 2x})}^{2} }$

$\tt\implies \: \boxed{ \pink{ \bf \: \dfrac{dy}{dx} = \dfrac{ - 8}{ {({e}^{2x} - {e}^{ - 2x})}^{2} } }}$