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(d/dx)log(cosx)=(d/dcosx)log(cosx)*(d/dx)cosx, first find dlog(cosx)/d(cosx)=f(x),let y=cos(x)
f(x)=limΔy→0(log(y+Δy)-log(y))/Δy=limΔy→0(log(1+ Δy/y)(1/Δy))=log(e(1/y))=1/y=1/cosx,d(cosx)/dx=
limΔx→0(cos(x+Δx)-cosx)/Δx =limΔx→0((cosx)cos(Δx)-(sinx)sin(Δx)-(cosx))/Δx =-sinx
therefore, answer is (1/cosx)(-sinx)=-tanx
use the chain rule i.e dy/dx=dy/dt * dt/dx .......dlog(cosx)/dx=dlog(cosx)/d(cosx) * d(cosx)/dx=1/(cosx) * -(sinx)=-(tanx)............similarly u can apply first principle to them using this rule................plz approve my answer if u like it
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