# If $y$ $=$ $\log_{e}{(\cos{\sqrt{x}})}$, then find $\dfrac{dy}{dx}$

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.

### Find the differentiation of the function

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.

### Try Chain Rule for differentiating the function

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.

#### Find $f'[{g(x)}]$

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 $g{(x)}$.

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

#### Find $g'{(x)}$

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 $\cos{(\sqrt{x})}$.

#### Substitute them in Chain Rule

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 $x$.

### Use Chain Rule one more time to complete differentiation of function

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

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.

#### Find $f'[{g(x)}]$

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

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

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

#### Find $g'{(x)}$

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

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 them in Chain Rule

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

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

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

### Finding the Derivative of the function y

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 $x$.

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

### Simplifying the differentiated function

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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