Question

Differentiate the following w.r.t.x: `(1 + sinx°)/(1 - sinx°)`

Answer 1

Let y = `(1 + sinx°)/(1 − sinx°)` 

y = `(1 + sin((πx)/180))/(1 − sin((πx)/180))          ...[∵ x° = ((pix)/180)^°]`

Differentiating w.r.t. x, we get,

`dy/dx = d/dx [(1 + sin((πx)/180))/(1 − sin((πx)/180))]`

`dy/dx = ([1 − sin((πx)/180)]. d/dx [1 + sin((πx)/180)] − [1 + sin((πx)/180)]. d/dx [1 − sin((πx)/180)])/[1 − sin((πx)/180)]^2`

`dy/dx = ([1 − sin((πx)/(180))].[0 + cos((πx)/(180)). d/dx ((πx)/(180)) - [1 + sin((πx)/(180))].[0 − cos((πx)/(180)). d/dx ((πx)/(180))]))/[1 − sin((πx)/180)]^2`

`dy/dx = ((1 − sinx°)[(cosx°) × π/(180) × 1] - (1 + sinx°)[(− cosx°) × π/(180) × 1])/(1 − sinx°)^2`

`dy/dx = (π/(180)cosx°(1 − sinx° +  1 + sinx°))/(1 - sinx°)^2`

`dy/dx = (πcosx°)/(90(1 − sinx°)^2`.

Answer 2

Convert the angle from degrees to radians

`1^\circ = pi/180 radians => x^\circ = (pix)/180 radians`

`y = (1+sin((pix)/180))/(1-sin((pix)/180))`

Differentiate using the Quotient Rule

`dy/dx = (v(du)/dx - u (dv)/dx)/v^2`

`u = 1 + sin ((pix)/180) => (du)/dx = cos ((pix)/180)*(pi/180)`   ...(by Chain Rule)

`v = 1 - sin ((pix)/180) => (dv)/dx = -cos ((pix)/180) * pi/180 `   ...(by Chain Rule)

`dy/dx = ([1-sin((pix)/180)] * [pi/180 cos ((pix)/180)] - [1+sin ((pix)/180)] * [-pi/180 cos ((pix)/180)])/([1 - sin ((pix)/180)]^2)`

Simplify the expression

`dy/dx = (pi/180 cos ((pix)/180) [(1 - sin ((pix)/180)) - (-1) (1 + sin ((pix)/180))])/[1-sin ((pix)/180)]^2`

`dy/dx = (pi/180 cos ((pix)/180) [1 - sin ((pix)/180) + 1 + sin  ((pix)/180)])/[1 - sin ((pix)/180)]^2`

`dy/dx = (pi/180 cos ((pix)/180) * [2])/ [1 - sin ((pix)/180)]^2`

`dy/dx = (pi/90 cos ((pix)/180))/[1 - sin ((pix)/180)]^2`

Convert back to degree notation

`dy/dx = pi/90 * cos x^\circ/(1 - sin x^\circ)^2`