How do you differentiate #y=csctheta(theta+cottheta)#?

Question

4652 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-22T14:00:50

#(dy)/(d theta)=-csctheta(thetacottheta+cot^2theta-1+csc^2theta)#

we will need to use the product rule

#d/(dx)(color(red)(u)color(blue)(v))=color(blue)(v)color(red)((du)/(dx))+color(red)(u)color(blue)((dv)/(dx))#

#y=csctheta(theta+cottheta)#

#(dy)/(d theta)=color(blue)((theta+cottheta))color(red)(d/(d theta)(csctheta))+color(red)(csctheta)color(blue)((dy)/(d theta)(theta+cottheta)#

#(dy)/(d theta)=(theta+cottheta)(-cscthetacottheta)+csctheta(1-csc^2theta)#

tiding up.

#(dy)/(d theta)=-csctheta(thetacottheta+cot^2theta-1+csc^2theta)#