Question

Express derivative of cot x in the form cosx.cosec x.?!​

Answer 1

Step-by-step explanation:

cotx=cosx/sinx

sinx = 1/cosecx

so

cotx=cosx×cosecx

Answer 2

Answer:

Step-by-step explanation:

We can write cotx=cosx cosecx

d/dx of (cosx cosecx)

= cosx(d/dx of cosec4x) + cosecx(d/dxof cosx) [d/dx of

cosecx= - cosec cotx]

= - cosx cosecx cotx + cosecx(-sinx) [d/dx of cosx =

-sinx]

= - cot^2x - 1

= - (cot^2x+1) [1+cot^2 x=cosec^2x]

= - cosec^2 x (ans)