
How do you find the derivative of $y = \ln (\cos (x))$ ?
Answer
545.4k+ views
Hint: In the given question there is a function inside another function. So, to find the derivative of a given function we need to make use of chain rule. Chin rule is nothing but first finding the derivative of the outer function and then the inner function.
Complete step-by-step answer:
In the given question there is a function inside another function. So, to find the derivative of a given function we need to make use of chain rule. Chin rule is nothing but first finding the derivative of the outer function and then the inner function.
The chain rule for differentiating the function inside function is given by:
$\dfrac{d}{{dx}}f[g(x)] = f'[g(x)].g'(x)$ here $f[g(x)]$ is nothing but the given function $y$.
By looking at the above chain rule equation we can relate the given function as $f[g(x)] = \ln (\cos x)$ and $g(x) = \cos x$ .
Now first we need to differentiate the function that is $f[g(x)] = \ln (\cos x)$ and then then inner function $g(x) = \cos x$ .
(the derivative of $\ln x = \dfrac{1}{x}$ , we make use of this while differentiating the outer function)
Therefore, on substituting the given function in the above chain rule formula, we get
$\dfrac{d}{{dx}}f[g(x)] = \dfrac{d}{{dx}}\ln (\cos x)$
On differentiating the outer function, we get
$ \Rightarrow \dfrac{d}{{dx}}f[g(x)] = \dfrac{1}{{\cos x}}.\dfrac{d}{{dx}}\cos x$
Now, we need to differentiate the inner function, that is $g(x) = \cos x$
(we know that the differentiation of $\cos x$ is $ - \sin x$)
$ \Rightarrow \dfrac{d}{{dx}}f[g(x)] = \dfrac{1}{{\cos x}}. - \sin x$
We can also write the above expression as
$ \Rightarrow \dfrac{d}{{dx}}f[g(x)] = - \dfrac{{\sin x}}{{\cos x}}$
The expression can be simplified further, so we get
$ \Rightarrow \dfrac{d}{{dx}}f[g(x)] = - \tan x$
Therefore the derivative of $y = \ln (\cos (x))$ is $ - \tan x$ .
Note: Chain rule has to be remembered if we have a function inside another function. The given problem can be solved in a shorter way just by differentiating the inner function and finally writing this differentiated inner function divided by the original inner function we get the required answer.
Complete step-by-step answer:
In the given question there is a function inside another function. So, to find the derivative of a given function we need to make use of chain rule. Chin rule is nothing but first finding the derivative of the outer function and then the inner function.
The chain rule for differentiating the function inside function is given by:
$\dfrac{d}{{dx}}f[g(x)] = f'[g(x)].g'(x)$ here $f[g(x)]$ is nothing but the given function $y$.
By looking at the above chain rule equation we can relate the given function as $f[g(x)] = \ln (\cos x)$ and $g(x) = \cos x$ .
Now first we need to differentiate the function that is $f[g(x)] = \ln (\cos x)$ and then then inner function $g(x) = \cos x$ .
(the derivative of $\ln x = \dfrac{1}{x}$ , we make use of this while differentiating the outer function)
Therefore, on substituting the given function in the above chain rule formula, we get
$\dfrac{d}{{dx}}f[g(x)] = \dfrac{d}{{dx}}\ln (\cos x)$
On differentiating the outer function, we get
$ \Rightarrow \dfrac{d}{{dx}}f[g(x)] = \dfrac{1}{{\cos x}}.\dfrac{d}{{dx}}\cos x$
Now, we need to differentiate the inner function, that is $g(x) = \cos x$
(we know that the differentiation of $\cos x$ is $ - \sin x$)
$ \Rightarrow \dfrac{d}{{dx}}f[g(x)] = \dfrac{1}{{\cos x}}. - \sin x$
We can also write the above expression as
$ \Rightarrow \dfrac{d}{{dx}}f[g(x)] = - \dfrac{{\sin x}}{{\cos x}}$
The expression can be simplified further, so we get
$ \Rightarrow \dfrac{d}{{dx}}f[g(x)] = - \tan x$
Therefore the derivative of $y = \ln (\cos (x))$ is $ - \tan x$ .
Note: Chain rule has to be remembered if we have a function inside another function. The given problem can be solved in a shorter way just by differentiating the inner function and finally writing this differentiated inner function divided by the original inner function we get the required answer.
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