Question

Find the derivative of `cscx` from first principles?

Answer 1

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

Explanation:

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`