Question

What is the derivative of this function `y= (1+sin x)/(1-sin x)`?

Answer 1

`dy/dx = ( 2cosx ) / (1-sinx)^2`

Explanation:

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=(1+sinx)/(1-sinx)` Then

`{ ("Let "u=1+sinx, => , (du)/dx=cosx), ("And "v=1-sinx, =>, (dv)/dx=-cosx ) :}`

`:. d/dx(u/v) = (v(du)/dx-u(dv)/dx)/v^2`
`:. dy/dx = ( (1-sinx)(cosx) - (1+sinx)(-cosx) ) / (1-sinx)^2`
`:. dy/dx = ( cosx-sinxcosx + cosx +sinxcosx ) / (1-sinx)^2`
`:. dy/dx = ( 2cosx ) / (1-sinx)^2`