Question

If y = sin (log (log 3x)), find
dy
dx​

Answer 1

given:

$y = \sin( log( log(3x) ) )$

$let \: log(3x) = t \\ on \: differentiating \: both \: sides \: w.r.to \: \: x,we \: get \\ \frac{1}{3x} = \frac{dt}{dx} \\ = > \frac{dt}{dx} = \frac{1}{3x} \: \: \: \: \: eqn(1) \\ now,\: \: \: y = \sin( log(t) ) \\ again, \: let \: log(t) = u \\ on \: differentiating \: both \: sides \: w.r.to \: \: t,we \: get \\ \frac{1}{t} = \frac{du}{dt} \\ = > \frac{du}{dt} = \frac{1}{t} \: \: \: \: \: \: \: eqn(2) \\ now, \: y = \sin(u) \\ differentiating \: w.r.to \: \: x,we \: get \\ \frac{dy}{dx} = \cos(u) \times \frac{du}{dx} \\ = > \frac{dy}{dx} = \cos(u) \times \frac{du}{dt} \times \frac{dt}{dx} \\ put \: the \: value \: of \: u \: and \: t \: from \: eqn(1) \: and \: eqn(2),, \\ = > \frac{dy}{dx} = \cos( log(t) ) \times \frac{1}{t} \times \frac{1}{3x} \\ = > \frac{dy}{dx} = \cos( log( log(3x) ) ) \times \frac{1}{ log(3x) } \times \frac{1}{3x} \\ hence, \: \frac{dy}{dx} = \frac{ \cos( log( log(3x) ) ) }{3x log(3x) }$