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Calculus1 24 Online
OpenStudy (anonymous):

use quotient rule to show d/dx(cscx)=-cscxcotx

OpenStudy (anonymous):

hi do you know whats identity of 1/cscx?

OpenStudy (anonymous):

no

OpenStudy (anonymous):

csc x=1/sin x do the derivative of this using the quotient rule d(csc x)dx=d(1/sin x)dx=?

OpenStudy (anonymous):

quotient rule says d(u/v)dx=(v du/dx - u dv/dx)/v^2

OpenStudy (anonymous):

u+1 and v=sin x \[d(\frac{ 1 }{ \sin x })/dx=\frac{ \sin x(d(1)/dx)- 1 d(\sin x/dx)}{ (\sin x)^2 }\] =\[\frac{ -\cos x }{ (\sin x)^2 }=\frac{ 1 }{ \sin x }*\frac{ \cos x }{ \sin x }=-\csc x \cot x\]

OpenStudy (anonymous):

that was u=1 and v=sin x

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