Differentiate #cos(x-2)/sinx#?

1 Answer
May 10, 2018

#f(x)=cos(x-2)/sinx=(cosxcos2-sinxsin2)/sinx#
#f(x)=cos2cotx-sin2#
#:.f'(x)=cos2(-csc^2x)-0=-cos2csc^2x=-cos2/sin^2x#
OR
Please see below.

Explanation:

We take,

#f(x)=(cos(x-2))/sinx#

Diff.w.r.t. #x#,#"using "color(blue)"Quotient Rule :"#

#color(blue)(d/(dx)(u*v)=(v*(du)/(dx)-u*(dv)/(dx))/v^2#

#f'(x)=(sinx*d/(dx)(cos(x-2))-cos(x-2)d/(dx)(sinx))/((sinx)^2)#

#f'(x)=(sinx(-sin(x-2))-cos(x-2)cosx)/sin^2x#

#f'(x)=-([cos(x-2)cosx+sin(x-2)sinx])/sin^2x#

#f'(x)=-[cos((x-2)-x)]/sin^2x=-cos(-2)/sin^2x#

#f'(x)=-cos2/sin^2x#