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Hint: The derivative of the function \[{{e}^{ax}}\] is given as \[\dfrac{d({{e}^{ax}})}{dx}=a.{{e}^{ax}}\] .

We are given \[y=\cos \text{ }hx=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}\], which is a hyperbolic function , and we need to find the derivative of the given function .

We will differentiate the given hyperbolic function with respect to \[x\].

On differentiating the given hyperbolic function with respect to \[x\], we get

\[\dfrac{dy}{dx}=\dfrac{d}{dx}\left( \cos \text{ }hx \right)\]

\[=\dfrac{d}{dx}\left( \dfrac{{{e}^{x}}+{{e}^{-x}}}{2} \right)\]

\[=\dfrac{d}{dx}(\dfrac{{{e}^{x}}}{2})+\dfrac{d}{dx}(\dfrac{{{e}^{-x}}}{2})..........\]equation\[(1)\].

Now , to find the derivative of the function , first we need to find the derivative of \[{{e}^{x}}\]and \[{{e}^{-x}}\] with respect to \[x\] .

We already know that the derivative of \[{{e}^{ax}}\]is \[\dfrac{d}{dx}{{e}^{ax}}=a.{{e}^{ax}}\]

So , the derivative of \[{{e}^{-x}}\] with respect to \[x\]can be calculated as \[\dfrac{d}{dx}{{e}^{-x}}=-1.{{e}^{-x}}=-{{e}^{-x}}\].

And , the derivative of \[{{e}^{x}}\] with respect to \[x\] can be calculated as \[\dfrac{d}{dx}{{e}^{x}}=1.{{e}^{x}}={{e}^{x}}\]

Now , to evaluate the derivative of the function , we will substitute the values of \[\dfrac{d}{dx}{{e}^{-x}}\] and \[\dfrac{d}{dx}{{e}^{x}}\] in equation\[(1)\].

On substituting the values of \[\dfrac{d}{dx}{{e}^{-x}}\] and \[\dfrac{d}{dx}{{e}^{x}}\] in equation\[(1)\], we get

\[\dfrac{dy}{dx}=\dfrac{{{e}^{x}}-{{e}^{-x}}}{2}\]

But we know that \[\dfrac{{{e}^{x}}-{{e}^{-x}}}{2}\] is the expansion of a hyperbolic function , \[\sinh x\].

So , we can write the value of the derivative of the function \[y\] as \[\dfrac{dy}{dx}=\sinh x\].

Hence , the value of the derivative of the hyperbolic function \[y=\cos \text{ }hx=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}\] is given as \[\dfrac{dy}{dx}=\sinh x\].

Note: Remember the expansion of \[\sin \text{ }hx\] is \[\dfrac{{{e}^{x}}-{{e}^{-x}}}{2}\]. Also remember that \[\dfrac{d}{dx}\left( \text{cos }hx \right)=\sin \text{ }hx\]and not \[-\sinh x\] . Students generally get confused and end up getting a wrong answer . Hence , such mistakes should be avoided .

We are given \[y=\cos \text{ }hx=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}\], which is a hyperbolic function , and we need to find the derivative of the given function .

We will differentiate the given hyperbolic function with respect to \[x\].

On differentiating the given hyperbolic function with respect to \[x\], we get

\[\dfrac{dy}{dx}=\dfrac{d}{dx}\left( \cos \text{ }hx \right)\]

\[=\dfrac{d}{dx}\left( \dfrac{{{e}^{x}}+{{e}^{-x}}}{2} \right)\]

\[=\dfrac{d}{dx}(\dfrac{{{e}^{x}}}{2})+\dfrac{d}{dx}(\dfrac{{{e}^{-x}}}{2})..........\]equation\[(1)\].

Now , to find the derivative of the function , first we need to find the derivative of \[{{e}^{x}}\]and \[{{e}^{-x}}\] with respect to \[x\] .

We already know that the derivative of \[{{e}^{ax}}\]is \[\dfrac{d}{dx}{{e}^{ax}}=a.{{e}^{ax}}\]

So , the derivative of \[{{e}^{-x}}\] with respect to \[x\]can be calculated as \[\dfrac{d}{dx}{{e}^{-x}}=-1.{{e}^{-x}}=-{{e}^{-x}}\].

And , the derivative of \[{{e}^{x}}\] with respect to \[x\] can be calculated as \[\dfrac{d}{dx}{{e}^{x}}=1.{{e}^{x}}={{e}^{x}}\]

Now , to evaluate the derivative of the function , we will substitute the values of \[\dfrac{d}{dx}{{e}^{-x}}\] and \[\dfrac{d}{dx}{{e}^{x}}\] in equation\[(1)\].

On substituting the values of \[\dfrac{d}{dx}{{e}^{-x}}\] and \[\dfrac{d}{dx}{{e}^{x}}\] in equation\[(1)\], we get

\[\dfrac{dy}{dx}=\dfrac{{{e}^{x}}-{{e}^{-x}}}{2}\]

But we know that \[\dfrac{{{e}^{x}}-{{e}^{-x}}}{2}\] is the expansion of a hyperbolic function , \[\sinh x\].

So , we can write the value of the derivative of the function \[y\] as \[\dfrac{dy}{dx}=\sinh x\].

Hence , the value of the derivative of the hyperbolic function \[y=\cos \text{ }hx=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}\] is given as \[\dfrac{dy}{dx}=\sinh x\].

Note: Remember the expansion of \[\sin \text{ }hx\] is \[\dfrac{{{e}^{x}}-{{e}^{-x}}}{2}\]. Also remember that \[\dfrac{d}{dx}\left( \text{cos }hx \right)=\sin \text{ }hx\]and not \[-\sinh x\] . Students generally get confused and end up getting a wrong answer . Hence , such mistakes should be avoided .

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