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

$y=cscx=1/sinx$

$u=1=>u'=0$

$v=sinx=>v'=cosx$

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

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

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

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

How do you find the derivative of cscxcscx?