Question

How do you differentiate `f(x)=sqrt((1-sinx)/(1+sinx))`?

Answer 1

`f'(x) = -( cosx) / ((1+sinx)^2) sqrt((1+sinx)/(1-sinx))`

Explanation:

Let `y=f(x)=sqrt((1-sinx)/(1+sinx)) => y^2 = (1-sinx)/(1+sinx)`

The LHS can be differentiated implicit, and the RHS can be differentiated using the quotient rule:

If you are studying maths, then you should learn the Quotient Rule for Differentiation, and practice how to use it:

`d/dx(u/v) = (v(du)/dx-u(dv)/dx)/v^2`, or less formally, `(u/v)' = (v(du)-u(dv))/v^2`

I was taught to remember the rule in word; " vdu minus udv all over v squared ". To help with the ordering I was taught to remember the acronym, VDU as in Visual Display Unit.

So with `y^2 = (1-sinx)/(1+sinx)` we have;

`2ydy/dx = ( (1+sinx)(d/dx(1-sinx)) - (1-sinx)(d/dx(1+sinx)) ) / (1+sinx)^2`

:. `2ydy/dx = ( (1+sinx)(-cosx) - (1-sinx)(cosx) ) / (1+sinx)^2`
:. `2ydy/dx = ( -cosx-sinxcosx-cosx+sinxcosx ) / (1+sinx)^2`
:. `2ydy/dx = ( -2cosx) / (1+sinx)^2`
:. `ydy/dx = -( cosx) / (1+sinx)^2`
:. `(sqrt((1-sinx)/(1+sinx)))dy/dx = -( cosx) / (1+sinx)^2`
:. `dy/dx = -( cosx) / ((1+sinx)^2sqrt((1-sinx)/(1+sinx)))`
:. `dy/dx = -( cosx) / ((1+sinx)^2) sqrt((1+sinx)/(1-sinx))`