if y= log{cot(pi/4+x/2)}. find dy/dx

`y=log{cot(pi/4+x/2)}`

Let `u=cot(pi/4+x/2)`

Apply the chain rule,

`dy/dx=d/(du)(log_10(u))d/dx(cot(pi/4+x/2))`

Now apply the common derivative,

`d/(du)(log_10(u))=1/(u ln(10))`

Now let's evaluate `d/dx(cot(pi/4+x/2))` by chain rule,

Let `u=pi/4+x/2`

`d/dx(cot(pi/4+x/2))=d/(du)(cot(u))d/dx(pi/4+x/2)`

Use the common derivative `d/(du)(cot(u))=-csc^2(u)`

and `d/dx(pi/4+x/2)=1/2`

`d/dx(cot(pi/4+x/2))=-csc^2(pi/4+x/2)*1/2`

and `dy/dx=(1/(u ln(10)))(-1/2csc^2(pi/4+x/2))`

Substitute back `u=cot(pi/4+x/2)`

`dy/dx=(1/(cot(pi/4+x/2) ln(10)))(-1/2csc^2(pi/4+x/2)`

`=-1/(2ln(10))(csc^2(pi/4+x/2))/cot(pi/4+x/2)`

We can further simplify as `csc^2(theta)=1/(sin^2(theta))`

and `cot(theta)=cos(theta)/(sin(theta))`

`:.dy/dx=-1/(2ln(10)) 1/(sin^2(pi/4+x/2))(sin(pi/4+x/2)/(cos(pi/4+x/2)))`

`=-1/(2ln(10)sin(pi/4+x/2)cos(pi/4+x/2))`

This can be simplified combining terms.

` -1/(ln(10)2sin(pi/4+x/2)cos(pi/4+x/2)) `

`=-1/(ln(10)sin(2(pi/4+x/2))) `

`=-1/(ln(10)sin(pi/2+x))=-1/(ln(10)cos(x)) `

This is the answer: `(-sec(x))/ln(10) `