could you please differentiate with respect to x and simplify? ln (1+cosx)/(1-cosx) thanks

Expert Answers
lemjay eNotes educator| Certified Educator


Before we differentiate, apply the quotient property of logarithm which is `ln(M/N)= lnM-lnN` .

`y=ln(1+cosx) - ln(1-cosx)`

Then, differentiate with respect to x.

`dy/dx=d/dx(ln(1+cosx) - ln(1-cosx))`

`dy/dx=d/dx(ln(1+cosx)) - d/dx(ln (1-cosx))`

To differentiate logarithm, apply the rule `(lnu)'=1/u*u'` .

`dy/dx=1/(1+cosx)*(-sinx) - 1/(1-cosx)*(sinx)`


Then, simplify.

`dy/dx= -(sinx (1-cosx))/((1-cosx)(1+cosx)) - (sinx(1+cosx))/((1-cosx)(1+cosx))`

`dy/dx=(-sinx+sinxcosx)/((1-cosx)(1+cosx)) - (sinx+sinxcosx)/((1-cosx)(1+cosx))`

`dy/dx= (-sinx +sinxcosx - sinx -sinxcosx)/((1-cosx)(1+cosx))`

`dy/dx= (-2sinx)/((1-cosx)(1+cosx))`


Applying the Pythagorean identity `sin^2x+cos^2=1` , the denominator becomes:


`dy/dx= (-2)/sinx`


Hence,  `dy/dx=-2/sinx` .