Question

(e^(x)+e^(-x))/(x^(2)+1)` differentiate​

Answer 1

Step-by-step explanation:

As per the information provided in the question, We have :

• y = (e^(x)+e^(-x))/(x^(2)+1)

In order to find the derivative of (e^(x)+e^(-x))/(x^(2)+1). We will use quotient rule.

It is given by –

$\longmapsto \rm \dfrac{d}{dx} y = \dfrac{d}{dx} \bigg ( \dfrac{u}{v} \bigg ) = \dfrac{v \dfrac{d}{dx} u - u \dfrac{d}{dx} v}{ {v}^{2} }$

By applying this We get,

${\longmapsto \rm \dfrac{ ({x}^{2} + 1) \dfrac{d}{dx} ( {e}^{x} + {e}^{ - x} ) - ( {e}^{x} + {e}^{ - x} )\dfrac{d}{dx} ( {x}^{2} + 1)}{ {({x}^{2} + 1)}^{2} } }$

${\longmapsto \rm { \dfrac{({x}^{2} + 1) ( {e}^{x} - {e}^{ - x} ) - ( {e}^{x} + {e}^{ - x} )\dfrac{d}{dx} ( {x}^{2} + 1)}{{({x}^{2} + 1)} ^{2} }} }$

${\longmapsto \rm \dfrac{({x}^{2} + 1) ( {e}^{x} - {e}^{ - x} ) - ( {e}^{x} + {e}^{ - x} )(2 \times {x}^{2 - 1} + 1)}{ {({x}^{2} + 1)}^{2} } }$

${\longmapsto \rm \dfrac{({x}^{2} + 1) ( {e}^{x} - {e}^{ - x} ) - ( {e}^{x} + {e}^{ - x} )(2 {x})}{ {({x}^{2} + 1)}^{2} } }$

${\longmapsto \rm \dfrac{({x}^{2} + 1) ( {e}^{x} - {e}^{ - x} ) - 2x( {e}^{x} + {e}^{ - x} )}{ {({x}^{2} + 1)}^{2} } }$

$\therefore \sf{ { Hence, \dfrac{({x}^{2} + 1) ( {e}^{x} - {e}^{ - x} ) - 2x( {e}^{x} + {e}^{ - x} )}{ {({x}^{2} + 1)}^{2} } \: is \: the \: derivative.}}$

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$\begin{gathered}\boxed{\begin{array}{cc}\bf{\dag}\:\:\underline{\text{Differentiation \:formulae:}}\\\\\bigstar\:\:\sf\dfrac{d}{dx} ( {log(x)}) =\dfrac{1}{x} \\\\\bigstar\:\:\sf\dfrac{d}{dx} ( {\sqrt{x}} )=\dfrac{1}{2 \sqrt{x}}\\\\\bigstar\:\:\sf\dfrac{d}{dx}({\sin(x)})=\cos(x)\\\\\bigstar\:\:\sf\dfrac{d}{dx}({\cos(x)})=-\sin(x)\\\\\bigstar\:\:\sf\dfrac{d}{dx}({\sin(x)})=\cos(x)\\\\\bigstar\:\:\sf\dfrac{d}{dx}({\tan(x)})={\sec}^{2}(x)\\\\\bigstar\:\:\sf\dfrac{d}{dx}({\cot(x)})=-{ \cosec}^{2}(x)\\\\\bigstar\:\:\sf\dfrac{d}{dx}({\sec(x)})=\sec(x)\times \tan(x)\\\\\bigstar\:\:\sf\dfrac{d}{dx}({\cosec(x)})= - \cosec(x)\times\cot(x)\end{array}}\end{gathered}$