Expert: Frederick Koh - 12/21/2011

Question
QUESTION: given that y= 1/cosx + 1/sinx show that dy/dx= sin^3x-cos^3x/(sin^2xcos^2x)

ANSWER: y= 1/cosx + 1/sinx =  sec x + cosec x

Differentiating both sides wrt x gives

dy/dx = (sec x) (tan x) -( cosec x)(cot x)
     = (1/cosx)(sinx/cosx) - (1/sinx) (cosx/sinx)
     = sin x/ cos^2  x - cos x/ sin^2 x
     = [sin x (sin^2 x) - cos x ( cos^2  x)]/(sin^2 x cos^2 x)  
     =  (sin^3 x - cos^3 x)/(sin^2 x cos^2 x)  (shown)

Hope this helps. Peace.

---------- FOLLOW-UP ----------

QUESTION: how do you find the coordinates of the point on the curve of y=1/cosx + 1/sinx in the range 0<x<pi where the gradient is zero?

Answer
If the gradient is 0, then dy/dx=0

(sin^3 x - cos^3 x)/(sin^2 x cos^2 x) =0

ie  sin^3 x - cos^3 x =0
   sin^3 x= cos^3 x
   tan^3 x= 1
   cube rooting both sides gives
   tan x =1

Hence x= arctan (1)= pi/4 (shown)

Hope this helps. Peace.