How do you differentiate #f(x) = [cot(x) + csc(x)]/[tan(x) - sin(x)]#?
1 Answer
Explanation:
First simplify
#f(x) = frac{cotx + cscx}{tanx - sinx}#
#= frac{cosx/sinx + 1/sinx}{sinx/cosx - sinx}#
#= frac{1/sinx(cosx + 1)}{sinx/cosx(1 - cosx)}#
#= frac{cosx}{sin^2x} * frac{(1 + cosx)}{(1 - cosx)}#
#= frac{cosx}{sin^2x} * frac{(1 + cosx)}{(1 - cosx)}* frac{(1 - cosx)}{(1 - cosx)}#
#= frac{cosx}{sin^2x} * frac{1 - cos^2x}{(1 - cosx)^2}#
#= frac{cosx}{sin^2x} * frac{sin^2x}{(1 - cosx)^2}#
#= frac{cosx}{(1 - cosx)^2}#
#= frac{cosx}{(2sin^2(x/2))^2}#
#= frac{cosx}{4sin^4(x/2)}#
Now we use the quotient rule to find the derivative of
#f'(x) = frac{"d"}{"d"x}(frac{cosx}{4sin^4(x/2)})#
#= frac{(4sin^4(x/2))frac{"d"}{"d"x}(cosx) - (cosx)frac{"d"}{"d"x}(4sin^4(x/2))}{(4sin^4(x/2))^2}#
#= frac{(4sin^4(x/2))(-sinx) - (cosx)(8sin^3(x/2)cos(x/2))}{16sin^8(x/2)}#
#= -frac{sin^2(x/2)sinx + cosx(2cos(x/2)sin(x/2))}{4sin^6(x/2)}#
#= -frac{sin^2(x/2)sinx + cosxsinx}{4sin^6(x/2)}#
#= frac{-sinx(sin^2(x/2) + cosx)}{4sin^6(x/2)}#
