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# Let $y=\cos \text{ }hx=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}$. Find the value of $\dfrac{dy}{dx}$.  Verified
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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 .