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

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

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)`

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

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))`

`dy/(dx)=(-2sinx)/(1-cos^2x)`

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

`dy/dx=(-2sinx)/(sin^2x)`

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

`dy/dx=-2/sinx`

Hence,  `dy/dx=-2/sinx` .

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