How do you find the derivative of #cscx#?

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Answer 1 · 2016-12-17T09:25:43

#(dy)/(dx)=-cotxcscx#

Rewrite #""cscx""# in terms of #""sinx""# and use the quotient rule

quotient rule #" "y=u/v=>(dy)/(dx)=(vu'-uv')/v^2#

#y=cscx=1/sinx#

#u=1=>u'=0#

#v=sinx=>v'=cosx#

#(dy)/(dx)=((sinx xx0)-(1xxcosx))/(sinx)^2#

#(dy)/(dx)=(0-cosx)/(sinx)^2#

#(dy)/(dx)=-cosx/(sinxsinx)=-cosx/sinx xx 1/sinx#

#(dy)/(dx)=-cotxcscx#