How do you use the quotient rule to differentiate #csc(t)/tan(t) #?

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Answer 1 · 2017-06-06T14:40:51

#d/dx(csct/tant)=-cscx (cot^2x + csc^2x)#

#d/dx((p(x))/(q(x)))=(q(x)p'(x)-p(x)q'(x))/([q(x)]^2)#

let #p(x)=csct#

#p'(x)=-csctcot t#

let #q(x)=tant#

#q'(x)=sec^2 t#

#d/dx(csct/tant)=(-csctcott(tant)-sec^2tcsct)/(tan^2t)#

#=-cscx (cot^2x + csc^2x)#