How do you differentiate #f(x)=x/sinx# using the quotient rule?

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Answer 1 · 2016-01-06T18:10:04

#f'(x) = (sinx-xcosx)/sin^2x#. For "how", see below.

Quotient rule: #d/dx(u/v) = (u'v-uv')/v^2#

In this function, we have #u=x# so #u'=1#

and #v=sinx#, so #v'=cosx#.

Apply the rule:

#f'(x) = ([1][sinx]-[x][cosx])/(sinx)^2#

# = (sinx-xcosx)/sin^2x#

This can, of course, be written in other ways using algebra and trigonometry.