y=log tanx show dy/dx=2cosec2x

Question

its from derivatives. class 12th

Answer 1

Given y=log{tanx}

from the definition of logarithm we get

tanx = 10^y

by appliying log(base)e or {ln}

ln{tanx} = y ln10

differentiating both sides

sec^2(x)/tanx = ln10 dy/dx

1/{tanxsec^2x} = ln10 dy/dx

1{sinxcosx} = ln10 dy/dx

= 2/{2sinxcosx} = ln10 dy/dx  --------------(multiplying 2 on the denominator and numerator of the L.H.S)

=2/sin(2x) = ln10 dy/dx ..................... = 2cosec(2x) =ln10 dy/dx

dy/dx ={2cosec(2x)}/ln10