How do you differentiate # g(x) = (cotx + cscx)(tanx - sinx) #?

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Answer 1 · 2015-12-06T16:14:11

#g'(x)=secxtanx+sinx#

First, simplify #g(x)#.

#g(x)=cotxtanx-cotxsinx+cscxtanx-cscxsinx#

#g(x)=1-cosx+secx-1#

#g(x)=secx-cosx#

Now, all we need to find #g'(x)# are the following identities:

#d/dx[secx]=secxtanx#

#d/dx[cosx]=-sinx#

Thus, #g'(x)=secxtanx+sinx#

This can also be written as: #g'(x)=sinx(sec^2x+1)#