Question

Differentiate logcosxw.r.to(cos(x))^1/2?

Answer 1

`(du)/(dv)=2/sqrt(cosx)`

Explanation:

Let ,`logcosx=log_e(cosx)=ln(cosx)`

We take,

`u=log(cosx) and v=(cosx)^(1/2)`

We have to find :`(du)/(dv)`

Now,

`(du)/(dx)=1/cosx xxd/(dx)(cosx)=1/cosx(-sinx)`

`(du)/(dx)=-sinx/cosx...to(1)`

And,

#(dv)/(dx)=1/2(cosx)^(1/2-1) d/(dx)(cosx)=1/(2sqrt(cosx))(- sinx)#

`(dv)/(dx)=-sinx/(2sqrt(cosx))...to(2)`

So, from `(1) and (2)`

#(du)/(dv)=((du)/(dx))/((dv)/(dx))=(-(sinx)/cosx)/(- sinx/(2sqrt(cosx)))=(2sqrt(cosx))/cosx#

`(du)/(dv)=2/sqrt(cosx)`

Answer 2

See below

Explanation:

Use differentials:

`(d ( logcos x))/(d( sqrt(cos x ) )`

`= (1/(cos x)* (- sin x) * dx)/(1/2 * 1/sqrt(cos x ) * (- sin x) * dx)`

`= 2/sqrt(cos x )`

If you're unfamiliar with differentials:

• let `z = sqrt( cos x)`

• `y = ln (cos x) = ln z^2`

`(d ( logcos x))/(d( sqrt(cos x ) )) equiv (dy)/(dz)`

`= d/(dz) (2 ln z) = 2/z`

`= 2/sqrt(cos x )`