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Question: If \(y=\dfrac{{{e}^{x}}+{{e}^{-x}}}{{{e}^{x}}-{{e}^{-x}}}\), then \[\dfrac{dy}{dx}\] is equal to ...

If y=ex+e−xex−e−xy=\dfrac{{{e}^{x}}+{{e}^{-x}}}{{{e}^{x}}-{{e}^{-x}}}, then dydx\dfrac{dy}{dx} is equal to
(a) sec⁡h2x\sec {{h}^{2}}x
(b) cos⁡ech2x\cos ec{{h}^{2}}x
(c) −sec⁡h2x-\sec {{h}^{2}}x
(d) −cos⁡ech2x-\cos ec{{h}^{2}}x

Explanation

Solution

Hint:Convert the given function to hyperbolic function by multiplying and dividing by 2 then differentiate accordingly.
Complete step-by-step answer:
Given,
y=ex+e−xex−e−xy=\dfrac{{{e}^{x}}+{{e}^{-x}}}{{{e}^{x}}-{{e}^{-x}}}
Dividing the numerator as well as denominator by 2, we get
y=ex+e−x2ex−e−x2…….(i)y=\dfrac{\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}}{\dfrac{{{e}^{x}}-{{e}^{-x}}}{2}}\ldots \ldots .\left( i \right)
Let’s convert this into hyperbolic functions.
Hyperbolic functions are exponential functions and have similar properties as the trigonometric functions. They are analogous to trigonometric functions. They are defined on hyperbola instead of circle.
We know,
sin⁡hx=ex−e−x2,cos⁡hx=ex+e−x2\sin hx=\dfrac{{{e}^{x}}-{{e}^{-x}}}{2},\cos hx=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}
So the equation (i) can be written as,
⇒y=cos⁡hxsin⁡hx\Rightarrow y=\dfrac{\cos hx}{\sin hx}
Differentiating with respect to x, we get
dydx=ddx(cos⁡hxsin⁡hx)\dfrac{dy}{dx}=\dfrac{d}{dx}\left( \dfrac{\cos hx}{\sin hx} \right)
Now we will apply the quotient rule, i.e., (u(x)v(x))′=u′(x).v(x)−u(x).v′(x)v(x)2{{\left( \dfrac{u\left( x \right)}{v\left( x \right)} \right)}^{'}}=\dfrac{{u}'\left( x \right).v\left( x \right)-u\left( x \right).{v}'\left( x \right)}{v{{\left( x \right)}^{2}}}, so
dydx=ddx(cos⁡hx)×(sin⁡hx)−(cos⁡hx)ddx(sin⁡hx)(sin⁡hx)2\dfrac{dy}{dx}=\dfrac{\dfrac{d}{dx}\left( \cos hx \right)\times \left( \sin hx \right)-\left( \cos hx \right)\dfrac{d}{dx}\left( \sin hx \right)}{{{\left( \sin hx \right)}^{2}}}
We know differentiation of sin⁡hx\sin hx and cos⁡hx\cos hx is cos⁡hx\cos hx and sin⁡hx\sin hx, so above equation becomes,
⇒dydx=sin⁡hx×sin⁡hx−cos⁡hxcos⁡hx(sin⁡hx)2\Rightarrow \dfrac{dy}{dx}=\dfrac{\sin hx\times \sin hx-\cos hx\cos hx}{{{\left( \sin hx \right)}^{2}}}
⇒dydx=(sin⁡hx)2−(cos⁡hx)2(sin⁡hx)2\Rightarrow \dfrac{dy}{dx}=\dfrac{{{\left( \sin hx \right)}^{2}}-{{\left( \cos hx \right)}^{2}}}{{{\left( \sin hx \right)}^{2}}}
In hyperbolic functions, (cosh2x−sinh⁡2x=1)(cos{{h}^{2}}x-{{\sinh }^{2}}x=1), so above equation become,
⇒dydx=−(cos⁡h2x−sinh⁡2x)(sin⁡hx)2\Rightarrow \dfrac{dy}{dx}=\dfrac{-\left( \cos {{h}^{2}}x-{{\sinh }^{2}}x \right)}{{{\left( \sin hx \right)}^{2}}}
⇒dydx=−1(sin⁡hx)2\Rightarrow \dfrac{dy}{dx}=-\dfrac{1}{{{\left( \sin hx \right)}^{2}}}
But we know, cosec hx=1sin⁡hxcosec\text{ }hx=\dfrac{1}{\sin hx}, so above equation becomes,
⇒dydx=−cosech2x\Rightarrow \dfrac{dy}{dx}=-cosec{{h}^{2}}x
Hence, the correct option for the given question is option (d).
Option ‘D’ is the right answer.

Note: Another way to solve this problem is first differentiating the given expression as it is and then converts the obtained answer into hyperbolic form.