Let \[y=\cos \text{ }hx=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}\]. Find the value of \[\dfrac{dy}{dx}\].
Answer
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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 .
Last updated date: 23rd Sep 2023
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