Question

If we have a function as \[{{\left( \cos x \right)}^{{{\left( \cos x \right)}^{{{\left( \cos x \right)}^{{{.....}^{\infty }}}}}}}}\], then \[\dfrac{dy}{dx}\] is equal to
(a) \[-\dfrac{{{y}^{2}}\tan x}{1-y\ln \left( \cos x \right)}\]
(b) \[\dfrac{{{y}^{2}}\tan x}{1+y\ln \left( \cos x \right)}\]
(c) \[\dfrac{{{y}^{2}}\tan x}{1-y\ln \left( \sin x \right)}\]
(d) \[\dfrac{{{y}^{2}}\sin x}{1+y\ln \left( \sin x \right)}\]

Answer 1

Hint: Take natural log, i.e. log to the base ‘e’, on both sides of the expression y. Write the expression as \[\ln y=y\ln \left( \cos x \right)\] by using the formula: - \[\ln {{a}^{m}}=m\ln a\]. Now, differentiate both sides of the function and use the formula: - \[\dfrac{d\ln y}{dx}=\dfrac{1}{y}\left( \dfrac{dy}{dx} \right)\] on the left-hand side. One the right-hand side apply product rule for differentiation given as: - \[\dfrac{d\left( u\times v \right)}{dx}=v\times \dfrac{du}{dx}+u\times \dfrac{dv}{dx}\], where u and v are the functions. Simplify the expression and find the value of \[\dfrac{dy}{dx}\].

Complete step-by-step solution
We have been provided with the expression: -
\[\Rightarrow y={{\left( \cos x \right)}^{{{\left( \cos x \right)}^{{{\left( \cos x \right)}^{{{.....}^{\infty }}}}}}}}\] - (1)
This is an infinite series, so on taking natural log (log to the base ‘e’) on both sides, we get,
\[\Rightarrow \ln y=\ln {{\left( \cos x \right)}^{{{\left( \cos x \right)}^{{{\left( \cos x \right)}^{{{.....}^{\infty }}}}}}}}\]
Applying the formula: - \[\ln {{a}^{m}}=m\ln a\], we get,
\[\Rightarrow \ln y={{\left( \cos x \right)}^{{{\left( \cos x \right)}^{{{\left( \cos x \right)}^{{{.....}^{\infty }}}}}}}}\ln \left( \cos x \right)\]
Using equation (1), we can write the above expression as: -
\[\Rightarrow \ln y=y\ln \left( \cos x \right)\]
Now, differentiating both sides with respect to x, we get,
\[\Rightarrow \dfrac{d\left[ \ln y \right]}{dx}=\dfrac{d\left( y\ln \left( \cos x \right) \right)}{dx}\]
In the R.H.S, we have a product of two functions y and \[\ln \left( \cos x \right)\]. Assuming them as u and v respectively, we will get the differential of the form: -
\[\Rightarrow \dfrac{d\left( \ln y \right)}{dx}=\dfrac{d\left( u\times v \right)}{dx}\]
Applying the product rule for differentiation given as: - \[\dfrac{d\left( u\times v \right)}{dx}=v\times \dfrac{du}{dx}+u\times \dfrac{dv}{dx}\], we get,
\[\Rightarrow \dfrac{1}{y}\left( \dfrac{dy}{dx} \right)=v\times \dfrac{du}{dx}+u\times \dfrac{dv}{dx}\]
Substituting the assumed values of u and v, we get,
\[\Rightarrow \dfrac{1}{y}\left( \dfrac{dy}{dx} \right)=\ln \left( \cos x \right)\times \dfrac{dy}{dx}+y\times \dfrac{d\ln \left( \cos x \right)}{dx}\]
Applying the chain rule for the term, \[\dfrac{d\ln \left( \cos x \right)}{dx}\], we have,
\[\begin{align}
  & \Rightarrow \dfrac{1}{y}\left( \dfrac{dy}{dx} \right)=\ln \left( \cos x \right)\times \dfrac{dy}{dx}+y\times \dfrac{d\ln \left( \cos x \right)}{d\cos x}\times \dfrac{d\cos x}{dx} \\
 & \Rightarrow \dfrac{1}{y}\left( \dfrac{dy}{dx} \right)=\dfrac{dy}{dx}\ln \left( \cos x \right)+y\times \dfrac{1}{\cos x}\times \left( -\sin x \right) \\
 & \Rightarrow \dfrac{1}{y}\left( \dfrac{dy}{dx} \right)=\dfrac{dy}{dx}\ln \left( \cos x \right)-y\left( \dfrac{\sin x}{\cos x} \right) \\
\end{align}\]
Applying the conversion, \[\dfrac{\sin \theta }{\cos \theta }=\tan \theta \], we get,
\[\begin{align}
  & \Rightarrow \dfrac{1}{y}\left( \dfrac{dy}{dx} \right)=\dfrac{dy}{dx}\ln \left( \cos x \right)-y\tan x \\
 & \Rightarrow {y}\tan x=\dfrac{dy}{dx}\ln \left( \cos x \right)-\dfrac{1}{y}\left( \dfrac{dy}{dx} \right) \\
\end{align}\]
Taking \[\left( \dfrac{dy}{dx} \right)\] common in the R.H.S, we get,
\[\Rightarrow {y}\tan x=\dfrac{dy}{dx}\left[ \ln \left( \cos x \right)-\dfrac{1}{y} \right]\]
\[\Rightarrow {y}\tan x=\dfrac{dy}{dx}\left[ \dfrac{y\ln \left( \cos x \right)-1}{y} \right]\]
By cross – multiplication, we get,
\[\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=\dfrac{{{y}^{2}}\tan x}{y\ln \left( \cos x \right)-1} \\
 & \Rightarrow \dfrac{dy}{dx}=-\dfrac{{{y}^{2}}\tan x}{1-y\ln \left( \cos x \right)} \\
\end{align}\]
Hence, option (a) is the correct answer.

Note: One must remember how to convert an infinite series into a simple form because we will not be able to solve the question without using the above process. You can also write the function as, \[y={{\left( \cos x \right)}^{y}}\] but then also when we take the log on both sides, we will get the same expression. You have to take all the terms of \[\dfrac{dy}{dx}\] at one side and then take \[\dfrac{dy}{dx}\] common from them because if any of the term containing \[\dfrac{dy}{dx}\] is left then we will get the answer in terms of \[\dfrac{dy}{dx}\], which will not be a simplified form.