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

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The denominator simplifies to:

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

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

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

So the answer simplifies to `(-sec(x))/ln(10) `

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Let `u=cot(pi/4+x/2)`

Apply the chain rule,


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`


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

and `d/dx(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)`


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)))`


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) `


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