find dy/dx if y = cos [logx]2

Given, y = coslog x2So, dydx=-sinlog x2×ddxlog x2=-sinlog x2×2log x2-1×ddxlogx=-2logx × sinlog x2 × 1x=-2 sin log x2× log xx

find dy/dx if y = cos [logx]²

Given, y = \cos {[\log x]}^{2} So, \frac{dy}{dx} = - \sin {[\log x]}^{2} \times \frac{d}{dx} {[\log x]}^{2} = - \sin {[\log x]}^{2} \times 2 {[\log x]}^{2 - 1} \times \frac{d}{dx} logx = - 2 logx \times \sin {[\log x]}^{2} \times \frac{1}{x} = - 2 \sin {[\log x]}^{2} \times \frac{\log x}{x}