The function $y$ is actually defined in terms of $x$. So, differentiate the function with respect to $x$ to get the derivative of $y$ in differential calculus.

Differentiate both sides with respect to $x$ to find $\dfrac{dy}{dx}$ in mathematics.

$\dfrac{dy}{dx}$ $\,=\,$ $\dfrac{d}{dx} \log_{e}{(\cos{\sqrt{x}})}$

There is a direct formula in differential calculus to differentiate the natural logarithmic function.

$\dfrac{d}{dx} \log_{e}{x}$ $\,=\,$ $\dfrac{1}{x}$

Actually, the function of $y$ is in different form though natural logarithmic function is involved in defining this function. So, the above differentiation rule cannot be applied to the function $y$ in this case because the function is formed by the composition of three different functions $\log_{e}{x}$, $\cos{x}$ and $\sqrt{x}$.

In this type of cases, the chain rule is used to find the derivative of the function, formed by the composition of two or more functions.

The chain rule can be written in mathematical form as follows.

$\dfrac{d}{dx} {f[{g(x)}]} \,=\, {f'[{g(x)}]}.{g'{(x)}}$

Take $f[{g(x)}]$ $\,=\,$ $\log_{e}{(\cos{\sqrt{x}})}$ and $g{(x)}$ $\,=\,$ $\cos{\sqrt{x}}$. Firstly, find $f'[{g(x)}]$ and $g'{(x)}$, and then substitute them in chain rule.

Differentiate the function $f[{g(x)}]$ with respect to $x$ to get $f'[{g(x)}]$.

$f'[{g(x)}]$ $\,=\,$ $\dfrac{d}{dx} {f[{g(x)}]}$ $\,=\,$ $\dfrac{d}{dx} \log_{e}{(\cos{\sqrt{x}})}$

Remember, just differentiate natural logarithmic function $\log_{e}{(\cos{\sqrt{x}})}$ by considering it as $\log_{e}{x}$ and don’t differentiate the function $\cos{\sqrt{x}}$ because the meaning of function $f'[{g(x)}]$ is the differentiation of function $f$ in terms of $g{(x)}$.

$\implies f'[{g(x)}]$ $\,=\,$ $\dfrac{1}{\cos{\sqrt{x}}}$

Similarly, differentiate the function $g{(x)}$ with respect to $x$ to get $g'{(x)}$.

$g'{(x)} \,=\, \dfrac{d}{dx} g{(x)}$ $\,=\,$ $\dfrac{d}{dx} \cos{(\sqrt{x})}$

The differentiation of $\cos{x}$ can be done by derivative of cos rule but there is no formula for derivative of $\cos{(\sqrt{x})}$.

Now, substitute the differentiated functions in chain rule formula.

$\dfrac{d}{dx} \log_{e}{(\cos{\sqrt{x}})}$ $\,=\,$ $\dfrac{1}{\cos{\sqrt{x}}}$ $\times$ $\dfrac{d}{dx} \cos{(\sqrt{x})}$

The differentiation of $\cos{(\sqrt{x})}$ is required to perform to complete the differentiation of the function $y$ but it cannot be performed directly because $\cos{(\sqrt{x})}$ is a composition of the functions $\cos{x}$ and $\sqrt{x}$. So, use chain rule one more time in order to find the derivative of function $\cos{(\sqrt{x})}$ with respect to $x$.

$\dfrac{d}{dx} {f[{g(x)}]}$ $\,=\,$ $\dfrac{d}{dx} \cos{(\sqrt{x})}$

Now, take $f[{g(x)}]$ $\,=\,$ $\cos{(\sqrt{x})}$ and $g{(x)} \,=\, \sqrt{x}$, then find $f'[{g(x)}]$ and $g'{(x)}$. After that replace differentiated functions $f'[{g(x)}]$ and $g'{(x)}$ in chain rule.

$f'[{g(x)}]$ $\,=\,$ $\dfrac{d}{dx} {f[{g(x)}]}$ $\,=\,$ $\dfrac{d}{dx} \cos{\sqrt{x}}$

Differentiate the function $\cos{(\sqrt{x})}$ with respect to $x$ by considering it as $\cos{x}$.

$\implies f'[{g(x)}]$ $\,=\,$ $-\sin{\sqrt{x}}$

$g'{(x)} \,=\, \dfrac{d}{dx} g{(x)}$ $\,=\,$ $\dfrac{d}{dx} \sqrt{x}$

Differentiate the square root of $x$ with respect to $x$ by using derivative of square root of $x$ formula.

$\implies g'{(x)} \,=\, \dfrac{1}{2\sqrt{x}}$

Substitute the functions in chain rule formula to get the differentiation of $\cos{(\sqrt{x})}$.

$\dfrac{d}{dx} \cos{(\sqrt{x})}$ $\,=\,$ $-\sin{\sqrt{x}}$ $\times$ $\dfrac{1}{2\sqrt{x}}$

$\implies \dfrac{d}{dx} \cos{(\sqrt{x})}$ $\,=\,$ $\dfrac{-\sin{\sqrt{x}}}{2\sqrt{x}}$

According to first time chain rule, the differentiation of the function $\log_{e}{(\cos{\sqrt{x}})}$ is calculated as follows.

$\dfrac{d}{dx} \log_{e}{(\cos{\sqrt{x}})}$ $\,=\,$ $\dfrac{1}{\cos{\sqrt{x}}}$ $\times$ $\dfrac{d}{dx} \cos{(\sqrt{x})}$

But the differentiation of $\cos{(\sqrt{x})}$ is also calculated as follows.

$\dfrac{d}{dx} \cos{(\sqrt{x})}$ $\,=\,$ $\dfrac{-\sin{\sqrt{x}}}{2\sqrt{x}}$

Therefore, replace the differentiation of $\cos{(\sqrt{x})}$ in the derivative of $\log_{e}{(\cos{\sqrt{x}})}$ with respect to $x$.

$\implies \dfrac{d}{dx} \log_{e}{(\cos{\sqrt{x}})}$ $\,=\,$ $\dfrac{1}{\cos{\sqrt{x}}}$ $\times$ $\dfrac{-\sin{\sqrt{x}}}{2\sqrt{x}}$

The differentiation of function $y$ is successfully completed and the differentiated function can be further simplified by the trigonometry.

$\implies \dfrac{d}{dx} \log_{e}{(\cos{\sqrt{x}})}$ $\,=\,$ $\dfrac{1}{2\sqrt{x}}$ $\times$ $\dfrac{-\sin{\sqrt{x}}}{\cos{\sqrt{x}}}$

$\implies \dfrac{d}{dx} \log_{e}{(\cos{\sqrt{x}})}$ $\,=\,$ $\dfrac{1}{2\sqrt{x}}$ $\times$ $-\tan{\sqrt{x}}$

$\,\,\, \therefore \,\,\,\,\,\, \dfrac{dy}{dx}$ $\,=\,$ $\dfrac{d}{dx} \log_{e}{(\cos{\sqrt{x}})}$ $\,=\,$ $\dfrac{-\tan{\sqrt{x}}}{2\sqrt{x}}$

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