Find the derivative of $cscx$ from first principles?

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Answer 1 · 2016-06-26T09:48:00

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

As $y=cscx=1/sinx$ , $y+deltay=1/sin(x+deltax)$

Hence $deltay=1/sin(x+deltax)-1/sinx$

= $(sinx-sin(x+deltax))/(sinxsin(x+deltax))$

= $-(sin(x+deltax)-sinx)/(sinxsin(x+deltax))$

= $-(2sin((x+deltax-x)/2)cos((x+deltax+x)/2))/(sinxsin(x+deltax))$

= $-(2sin((deltax)/2)cos(x+(deltax)/2))/(sinxsin(x+deltax))$ and

$(deltay)/(deltax)=-(2sin((deltax)/2)cos(x+(deltax)/2))/(deltaxsinxsin(x+deltax))$

Hence $(dy)/(dx)=Lt_(deltax->0)(deltay)/(deltax)$

= $Lt_(deltax->0)-(2sin((deltax)/2)cos(x+(deltax)/2))/(deltaxsinxsin(x+deltax))$

= $Lt_(deltax->0)(-sin((deltax)/2)/((deltax)/2))xxLt_(deltax->0)[(cos(x+(deltax)/2))/(deltaxsinxsin(x+deltax))]$

= $-1xxcosx/(sin^2x)=-1/(sinx)xxcosx/sinx=-cscxcotx$

Find the derivative of cscxcscx from first principles?