How do you differentiate #(x)/ (1-cosx)# using the quotient rule?

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Answer 1 · 2016-01-26T19:13:44

#(1-cosx-xsinx)/(1-cosx)^2#

The quotient rule states that

#d/dx[(f(x))/(g(x))]=(f'(x)g(x)-g'(x)f(x))/[g(x)]^2#

Applying this to #x/(1-cosx)#, we see that its derivative is equal to

#((1-cosx)d/dx[x]-xd/dx[1-cosx])/(1-cosx)^2#

We can find each of the internal derivatives and then plug them back in:

#d/dx[x]=1#

This one is a little trickier. Since #1# is a constant, all we are really finding is the derivative of #-cosx#, which is positive #sinx#.

#d/dx[1-cosx]=sinx#

Plugging it all back in:

#=(1-cosx-xsinx)/(1-cosx)^2#