Question

differentiate
log(sinx) w.r.t cotx​

Answer 1

Step-by-step explanation:

let us consider ,

U = log(sinx)

v = cotx

so,

du/dv= (du/dx )/ ( dv / dx)

✔so first differentiate U = log(sinx)

wrt x

du/dx = d/dx (log ( sinx) )

= (1 / sinx ) cosx

=( cosx / sinx)

✔Now, differentiate

v = cotx

dv / dx = - cosec^2 x

therefor

✔ du/dv= (du/dx )/ ( dv / dx)

= (( cosx / sinx) ) /(- cosec^2 x)

= ( cos x / - (sinx. cosec^2 x))

= ( - ( sin^2x . cosx) / sinx))

= - sinx. cosx