3(1-sinx) / (2cosx)
I was using difference q and my set-up looks like this: 3(cosx)(2cosx)-3(1-sinx)(-2sinx)/(2cosx)…
What is wrong with it? Thanks in advance.
I was using difference q and my set-up looks like this: 3(cosx)(2cosx)-3(1-sinx)(-2sinx)/(2cosx)…
What is wrong with it? Thanks in advance.
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I think you hve siGn problem. d/dx(1–sin x) = – cos x
It may be wise to factor the constants off to the left prior to differentiating using the quotient rule.
Or, convert to other functions before differentiating, like so:
3... 1 – sinx .......3....... 1......... sin x
-- • ------------ = --- • (--------- – --------- ) =
2...... cos x.........2..... cos x .... cos x
(3/2)[ sec x – tan x). Now differentiate!
Diferentiating we get (3/2)[sec x tan x – sec² x] =
(3/2) sec x ( tan x – sec x )
It may be wise to factor the constants off to the left prior to differentiating using the quotient rule.
Or, convert to other functions before differentiating, like so:
3... 1 – sinx .......3....... 1......... sin x
-- • ------------ = --- • (--------- – --------- ) =
2...... cos x.........2..... cos x .... cos x
(3/2)[ sec x – tan x). Now differentiate!
Diferentiating we get (3/2)[sec x tan x – sec² x] =
(3/2) sec x ( tan x – sec x )