Question

Y=log (logx) differentiate

Answer 1

Question :-

y = log ( log x) differentiate it .

Answer :-

$\: \: \: \: \: \: \: \: \: \: \: \to \boxed{\sf \: \frac{dy}{dx} = \frac{1}{x \: \log(x)} } \:$

Solution :-

→ we have ,

$\to \sf \: y = log \{log(x) \} \\ \\ \bf differentiate \: with \: respect \: to \: x \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \to \: \frac{dy}{dx} = log \{log(x) \\ \\ \boxed{ \sf\because \: \frac{d}{dx} \log x = \frac{1}{x} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \to \sf \: \frac{dy}{dx} = \frac{1}{ \log(x)} \frac{d}{dx} \{log(x) \} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \:\to \sf \frac{dy}{dx} = \frac{1}{ \log(x)} \times \frac{1}{x} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \to \boxed{\sf \: \frac{dy}{dx} = \frac{1}{x \: \log(x)} }$

Some Differential formula :-

$\star \sf \: \frac{d}{dx} {e}^{x} = {e}^{x} \\ \\ \star \sf \: \frac{d}{dx} {a}^{x} = {a}^{x} log_{e}(a) \\ \\ \star \sf \: \frac{d}{dx} \tan x = { \sec}^{2} x \\ \\ \star \sf \: \frac{dy}{dx} \: \sec x = \sec x \: . \tan x \\ \\ \star \sf \: \frac{d}{dx} { \tan}^{ - 1} x = \frac{1}{1 + {x}^{2} } \\ \\ \star \sf \: \frac{d}{dx} { \sin}^{ - 1} x = \frac{1}{ \sqrt{1 - {x}^{2} } } \\ \\ \star \sf \: \frac{d}{dx} \: { \sec}^{ - 1} x = \frac{1}{x \sqrt{ {x}^{2} - 1} }$