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

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,

& \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{aligned}$$ Applying the conversion, $$\dfrac{\sin \theta }{\cos \theta }=\tan \theta $$, we get, $$\begin{aligned} & \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{aligned}$$ Taking $$\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{aligned} & \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{aligned}$$ **Hence, option (a) is the correct answer.** **Note:** One must remember how to convert an infinite series into a simple form because we will not be able to solve the question without using the above process. You can also write the function as, $$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.