Question

Find the derivative of the below function w.r.t. x

$\boxed{y = {cos \: x}^{cosx^{cos \: x^{... \infty} } } }$​

Answer 1

Given function is,

$\small\rm \longrightarrow y = {cos \: x}^{cosx^{cos \: x^{... \infty}}} \dots \dots(1.)$

$\small\rm \longrightarrow y = {cos \: x}^{ \textbf{ \textsf{y}}} \: \: \left \{ using \: equation \: 1\right \}$

Taking ln ( log with natural base ) both sides,

$\small\rm \longrightarrow \ln (y) = \ln({ cos \: x}^{y})$

Apply property of log,

• $\boxed{ \sf \log {x}^{y} = y \log x}$

$\small\rm \longrightarrow \ln (y) = y\ln({ cos \: x})$

Now differentiating both sides w.r.t. x, we get:

$\small\rm \longrightarrow \dfrac{d}{dx} (\ln y) = \dfrac{d}{dx} (y\ln{ cos \: x})$

Using differentiation formulas:-

• $\boxed{\sf \dfrac{d}{dx} (\ln x) = \dfrac{1}{x} }$

• $\boxed {\sf \dfrac{d}{dx}f(x).g(x) =f'(x).g(x) + f(x).g'(x) }$

• $\boxed {\sf \dfrac{d}{dx}(cos\,x=-sin\, x }$

$\small\rm \longrightarrow \dfrac{1}{y} .\dfrac{dy}{dx} = \dfrac{dy}{dx}. \ln(cos \: x) + y. \dfrac{d}{dx} (\ln{ cos \: x})$

$\small\rm \longrightarrow \dfrac{1}{y} .\dfrac{dy}{dx} = \dfrac{dy}{dx}. \ln(cos \: x) + y. \dfrac{1}{ cos \: x}( - sin \: x)$

$\small\rm \longrightarrow \dfrac{1}{y} .\dfrac{dy}{dx} - \dfrac{dy}{dx}. \ln(cos \: x) = y \times \dfrac{ - sin \: x}{ cos \: x}$

$\small\rm \longrightarrow \dfrac{dy}{dx} \left( \dfrac{1}{y} - \ln cos \: x \right) = - y \:tan \: x$

$\small\rm \longrightarrow \dfrac{dy}{dx} = \dfrac{ - y \:tan \: x}{ \left( \frac{1}{y} - \ln cos \: x \right)}$

$\small\rm \longrightarrow \dfrac{dy}{dx} = \dfrac{ - y \:tan \: x}{ \left( \frac{1 - y \ln \: cos \: x}{y} \right)}$

$\small\rm \longrightarrow \dfrac{dy}{dx} = \dfrac{ - {y}^{2} \:tan \: x}{1 - y \ln \:cos \: x}$

$\small\rm \longrightarrow \dfrac{dy}{dx} = \dfrac{ - {y}^{2} \:tan \: x}{1 - y \ln( \:cos \: x)}$

Therefore,

$\purple{\boxed{ \small\sf \implies \dfrac{d}{dx} \left({cos \: x}^{cosx^{cos \: x^{... \infty}}} \right) = \dfrac{ - {y}^{2} \:tan \: x}{1 - y \ln( \:cos \: x)}}}$